A

Woken up by the alarm clock Igor the financial analyst hurried up to the work. He ate his breakfast and sat in his car. Sadly, when he opened his GPS navigator, he found that some of the roads in Bankopolis, the city where he lives, are closed due to road works. Moreover, Igor has some problems with the steering wheel, so he can make no more than two turns on his way to his office in bank.

Bankopolis looks like a grid of n rows and m columns. Igor should find a way from his home to the bank that has no more than two turns and doesn't contain cells with road works, or determine that it is impossible and he should work from home. A turn is a change in movement direction. Igor's car can only move to the left, to the right, upwards and downwards. Initially Igor can choose any direction. Igor is still sleepy, so you should help him.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 1000) — the number of rows and the number of columns in the grid.

Each of the next n lines contains m characters denoting the corresponding row of the grid. The following characters can occur:

• "." — an empty cell;
• "*" — a cell with road works;
• "S" — the cell where Igor's home is located;
• "T" — the cell where Igor's office is located.

It is guaranteed that "S" and "T" appear exactly once each.

Output

In the only line print "YES" if there is a path between Igor's home and Igor's office with no more than two turns, and "NO" otherwise.

Example

Input

5 5..S..****.T....****......

Output

YES

Input

5 5S....****.......****..T..

Output

NO

Note

The first sample is shown on the following picture:

In the second sample it is impossible to reach Igor's office using less that 4turns, thus there exists no path using no more than 2 turns. The path using exactly 4 turns is shown on this picture.

题意：给出S和T表示起点和终点。‘.'表示路，‘*'表示墙，问你是否能找出一条由S通往T的路，这条路的拐弯次数不能超过两次。

思路：很明显的Dfs，横向扫一下，竖向扫一下，搜到墙或者转弯次数超过2回朔就好，关键是看你转弯统计是否做得巧妙，话不多说，看代码。

#include<bits/stdc++.h>using namespace std;#define maxn 1005int n,m,ch,t,visit[maxn][maxn][5][5];char str[maxn][maxn];void dfs(int x,int y,int c,int t){     if(x<1||y<1||x>n||y>m) return;     if(t>2||str[x][y]=='*'||visit[x][y][c][t]||ch) return;      visit[x][y][c][t]=1;     if(str[x][y]=='T') {ch=1; return;}    if(!c) dfs(x-1,y,c,t),dfs(x+1,y,c,t);    else dfs(x,y-1,c,t),dfs(x,y+1,c,t);    dfs(x,y,!c,t+1);}int main(){    int x,y;    ch=0;    memset(visit,0,sizeof(0));    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) {cin>>str[i][j];if(str[i][j]=='S')x=i,y=j;}    dfs(x,y,0,0);dfs(x,y,1,0);    if(ch)cout<<"YES"<<endl;    else cout<<"NO"<<endl;    return 0;}