699B - One Bomb

B. One Bomb

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a description of a depot. It is a rectangular checkered field of n × m size. Each cell in a field can be empty (".") or it can be occupied by a wall ("*").

You have one bomb. If you lay the bomb at the cell (x, y), then after triggering it will wipe out all walls in the row xand all walls in the column y.

You are to determine if it is possible to wipe out all walls in the depot by placing and triggering exactly one bomb. The bomb can be laid both in an empty cell or in a cell occupied by a wall.

Input

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

The next n lines contain m symbols "." and "*" each — the description of the field. j-th symbol in i-th of them stands for cell (i, j). If the symbol is equal to ".", then the corresponding cell is empty, otherwise it equals "*" and the corresponding cell is occupied by a wall.

Output

If it is impossible to wipe out all walls by placing and triggering exactly one bomb, then print "NO" in the first line (without quotes).

Otherwise print "YES" (without quotes) in the first line and two integers in the second line — the coordinates of the cell at which the bomb should be laid. If there are multiple answers, print any of them.

Examples

input

3 4.*.......*..

output

YES1 2

input

3 3..*.*.*..

output

NO

input

6 5..*....*..*****..*....*....*..

output

YES3 3

给出n*m的一张图，‘*’ 代表墙，‘.’ 代表空地，现在有一个炸弹，如果把炸弹在某一个位置使用，可以炸掉这个位置所在的行和列的所有的墙，问是否能在某个位置使用，能使得地图上所有的墙都被破坏掉（最终只剩下空地）

先统计好每行和每列都有多少个炸弹，然后暴力枚举每一个位置，只要这个位置对应的行和列的炸弹数和总共的炸弹数相等，那么就输出这个位置，另外注意如果某个位置有墙的话，可能会多统计一次，需要讨论

刚开始感觉很简单，做着做着脑子混乱了，不知道怎么下手了.....后来直接暴力，水掉了...第一次打cf，惨淡收场....

/*http://blog.csdn.net/liuke19950717*/#include<cstdio>const int maxn=1005;char map[maxn][maxn];int n,m,x[maxn],y[maxn];void slove(int cnt){for(int i=0;i<n;++i){for(int j=0;j<m;++j){if(map[i][j]=='.'&&x[i]+y[j]==cnt||map[i][j]=='*'&&x[i]+y[j]==cnt+1){printf("YES\n%d %d\n",i+1,j+1);return;}}}printf("NO\n");}int main(){int cnt=0;scanf("%d%d",&n,&m);for(int i=0;i<n;++i){scanf("%s",map[i]);}for(int i=0;i<n;++i){int tp=0;for(int j=0;j<m;++j){if(map[i][j]=='*'){++tp;++cnt;}}x[i]=tp;}for(int i=0;i<m;++i){int tp=0;for(int j=0;j<n;++j){if(map[j][i]=='*'){++tp;}}y[i]=tp;}slove(cnt);return 0;}