//水题一道
#include<bits/stdc++.h>
#define MAXN 25
#define p 100007
#define int long long
using namespace std;
int n,m,K,now,pre;
char c[MAXN][MAXN];
int head[p+3],nxt[1<<24];
int f[2][1<<24],fst[2][3][1<<24];
int cnt[2];
void insert(int state1,int state2,int state3,int val)
{
	int ha=((((1ll*state1<<16)|state2)<<4)|state3)%p+1;//哈希
	for(int i=head[ha];i;i=nxt[i])
		if(fst[now][0][i]==state1&&fst[now][1][i]==state2&&fst[now][2][i]==state3)
		{
			f[now][i]=max(f[now][i],val);
			return;
		}
	f[now][++cnt[now]]=val;
	fst[now][0][cnt[now]]=state1;
	fst[now][1][cnt[now]]=state2;
	fst[now][2][cnt[now]]=state3;//三个状态
	nxt[cnt[now]]=head[ha];
	head[ha]=cnt[now];
}//hash表
int left(int state,int pos,int val)
{//向左寻找左括号
	int tot=1;
	while(true)
	{
		int pl=(state>>(pos<<1))&3;
		if(pl==2)tot++;if(pl==1) tot--;
		if(!tot)return state^(pl<<(pos<<1))^(val<<(pos<<1));
		pos--;
	}
}
int right(int state,int pos,int val)
{//向右寻找右括号
	int tot=1;
	while(true)
	{
		int pl=(state>>(pos<<1))&3;
		if(pl==1)tot++;if(pl==2) tot--;
		if(!tot)return state^(pl<<(pos<<1))^(val<<(pos<<1));
		pos++;
	}
}
char tmp[MAXN][MAXN];
signed main()
{
	scanf("%lld%lld%lld",&n,&m,&K);
	for(int i=0;i<n;i++)scanf("%s",c[i]);
	if(n<m)
	{//n<m的时候把地图转90度
		memcpy(tmp,c,sizeof(c));
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
				c[j][i]=tmp[i][j];
		swap(n,m);
	}
	cnt[0]=1;
	for(int i=0;i<n;i++)
	{
		for(int j=1;j<=cnt[now];j++)
			fst[now][0][j]<<=2;
		for(int j=1;j<=cnt[now];j++)
			fst[now][1][j]=(fst[now][1][j]^(((fst[now][1][j]>>(m+1<<1))&3)<<(m+1<<1)))<<2;//更新上一行的状态
		for(int j=0;j<m;j++)
		{
			pre=now;now^=1;cnt[now]=0;
			memset(head,0,sizeof(head));//注意初始化
			for(int k=1;k<=cnt[pre];k++)
			{
				int sta=fst[pre][0][k],stb=fst[pre][1][k],stc=fst[pre][2][k];
				int val=f[pre][k];
				if(sta>=(1<<(m+1<<1)))continue;
				int pl1=(sta>>(j<<1))&3;//right
				int pl2=(sta>>(j+1<<1))&3;//down
				int r=(stb>>(j<<1))&3;//右状态
				int ur=(stb>>(j+1<<1))&3;//右上状态
				int u=(stb>>(j+2<<1))&3;//上状态
				int ul=(stb>>(j+3<<1))&3;//左上状态
				int now=(sta^(pl1<<(j<<1))^(pl2<<(j+1<<1)));
				if((!pl1&&r==1)||(!pl2&&u==1))//1
				{
					if(pl1||pl2)continue;
					insert(now,stb^(ur<<(j+1<<1)),stc,val);//放障碍
					if(stc<K&&c[i][j]=='.')
						insert(now,stb^(ur<<(j+1<<1))^(3<<(j+1<<1)),stc+1,val+(r==1)+(ur==1)+(u==1)+(ul==1));//放炮台
				}
				else if(c[i][j]=='#'){if(!pl1&&!pl2)insert(now,stb^(ur<<(j+1<<1)),stc,val);}//2
				else if(c[i][j]=='.')//3
				{
					if(!pl1&&!pl2)//1)
					{
						if(stc<K)
							insert(now,stb^(ur<<(j+1<<1))^(3<<(j+1<<1)),stc+1,
								val+(r==1)+(ur==1)+(u==1)+(ul==1));//放炮台
						insert(now^(1<<(j<<1))^(2<<(j+1<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));//走
						insert(now,stb^(ur<<(j+1<<1)),stc,val);//不走
					}
					else if(!pl1&&pl2)//2）
					{
						insert(now^(pl2<<(j<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
						insert(now^(pl2<<(j+1<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					}
					else if(pl1&&!pl2)//2）
					{
						insert(now^(pl1<<(j<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
						insert(now^(pl1<<(j+1<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					}
					else if(pl1==1&&pl2==1)//3）
						insert(right(now,j,1),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==2&&pl2==2)//4）
						insert(left(now,j,2),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==2&&pl2==1)//6）
						insert(now,stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==1&&pl2==3)//7）
						insert(right(now,j,3),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==2&&pl2==3)//8）
						insert(left(now,j,3),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==3&&pl2==1)//7）
						insert(right(now,j,3),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==3&&pl2==2)//8）
						insert(left(now,j,3),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==3&&pl2==3)//9）
						insert(now,stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
				}
				else if(c[i][j]=='S'||c[i][j]=='T')//4
				{
					if(c[i][j]=='S'&&!pl1&&!pl2)//1）
					{
						insert(now^(3<<(j<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
						insert(now^(3<<(j+1<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					}
					else if(c[i][j]=='T'&&!pl1&&!pl2)//1）
					{
						insert(now^(3<<(j<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
						insert(now^(3<<(j+1<<1)),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					}
					else if(!pl1&&pl2==1)//2）
						insert(right(now,j,3),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(!pl1&&pl2==2)//3）
						insert(left(now,j,3),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==1&&!pl2)//2）
						insert(right(now,j,3),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==2&&!pl2)//3）
						insert(left(now,j,3),stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(pl1==3&&!pl2)//4）
						insert(now,stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
					else if(!pl1&&pl2==3)//4）
						insert(now,stb^(ur<<(j+1<<1))^(1<<(j+1<<1)),stc,
							val+(r==3)+(ur==3)+(u==3)+(ul==3));
				}
			}
		}
	}
	int ans=0;
	for(int i=1;i<=cnt[now];i++)//统计答案
		if(!fst[now][0][i])
			ans=max(ans,f[now][i]);
	printf("%lld",ans);
	return 0;
}