codeforces 699B One Bomb

Description

You are given a description of a depot. It is a rectangular checkered field ofn × 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 rowx and all walls in the columny.

You are to determine if it is possible to wipe out all walls in the depot by placing and triggeringexactly 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 andm (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 ini-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

一道cf的B题，大概搞了两个小时。。

思路就是统计每一行每一列墙壁的数量r[I],c[I],枚举每一个元素，看r[I]+c[j]-a[I][j]是否等于墙壁总数。

#include<stdio.h>#include<string.h>#include<ctype.h>#include<iostream>#include<vector>#include<sstream>#include<queue>#include<limits.h>#include<set>#include<math.h>#include<algorithm>using namespace std;int n,m,index_b=0;int r[1010],c[1010];int a[1010][1010];int main(void){    cin>>n>>m;    int num=0,flag=0;    string temp;    char tempp;    memset(r,0,sizeof(r));    memset(c,0,sizeof(c));    for(int i=1;i<=n;i++)    {        cin>>temp;        stringstream ss(temp);        for(int j=1;j<=m;j++)        {            ss>>tempp;            if(tempp=='*')            {                a[i][j]=1;                r[i]++;                c[j]++;                num++;            }            else                a[i][j]=0;        }    }    int ans_x,ans_y;    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)    {        if(r[i]+c[j]-a[i][j]==num)        {            flag=1;            ans_x=i;            ans_y=j;        }    }    if(!flag)        printf("NO");    else        printf("YES\n%d %d",ans_x,ans_y);    return 0;}