One Bomb

Description

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 x and 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.

Sample Input

Input

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

Output

YES1 2

Input

3 3..*.*.*..

Output

NO

Input

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

Output

YES3 3

题意；看是否存在一个点放炸弹，使的所有行和列的墙都被炸掉；

思路：

在输入时保存下每行每列的*的个数，后来逐个遍历，去看是否每行每列的个数与走的个数相同（如果遍历的这个符号是*,要记得减去1）

1；

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char map[1100][1100];int row[1100],col[1100];int main(){int n,m,i,j;while(scanf("%d%d",&n,&m)!=EOF){int ant=0;memset(row,0,sizeof(row));memset(col,0,sizeof(col));for(i=1;i<=n;i++)scanf("%s",map[i]);for(i=1;i<=n;i++){for(j=0;j<m;j++){if(map[i][j]=='*'){row[i]++;col[j]++;ant++;}}}int ok=0;for(i=1;i<=n;i++){for(j=0;j<m;j++)if(map[i][j]=='*'){if(row[i]+col[j]-1==ant){printf("YES\n%d %d\n",i,j+1);return 0;}}else{if(row[i]+col[j]==ant){printf("YES\n%d %d\n",i,j+1);return 0;}}}if(!ok)printf("NO\n");return 0;}}

2

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char map[1100][1100];int row[1100],col[1100];int main(){int n,m,i,j;while(scanf("%d%d",&n,&m)!=EOF){int ant=0,flag=0,x,y;memset(row,0,sizeof(row));memset(col,0,sizeof(col));for(i=0;i<n;i++)scanf("%s",map[i]);for(i=0;i<n;i++){for(j=0;j<m;j++){if(map[i][j]=='*'){row[i]++;col[j]++;ant++;}}}for(i=0;i<n;i++){for(j=0;j<m;j++){int s=row[i]+col[j];if(map[i][j]=='*')s--;if(s==ant){x=i;y=j;    flag=1;}}}  if(flag)printf("YES\n%d %d\n",x+1,y+1);elseprintf("NO\n");}return 0;}