Codeforces Round #363 (Div. 2) B. One Bomb（思维题目）

B. One Bomb

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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.

Examples

input

3 4

.*..

….

.*..

output

YES

1 2

input

3 3

..*

.*.

*..

output

NO

input

6 5

..*..

..*..

..*..

..*..

..*..

output

YES

3 3

说起来这个题也不算太难。以前见到过类似的，直接记录某一行和某一列的墙的数目。

然后暴力枚举每一个点，如果该点是墙，若该点所处的行和列的墙的总数为总墙数+1，则可以把所有的墙都炸掉，如果该点是平地，若该点所处的行和列的总数为总墙数，则可以把所有的墙都炸掉。

下面是AC代码：

#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;

char a[1005][1005];
int c[1005],d[1005];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int sum=0;
for(int i=0;i<n;i++)
{
scanf("%s",&a[i]);
for(int j=0;j<m;j++)
{
if(a[i][j]=='*')
{
sum++;
}
}
}
memset(c,0,sizeof(c));
memset(d,0,sizeof(d));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(a[i][j]=='*')
{
c[i]++;
d[j]++;
}
}
}
int flag=0,sx,sy;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(a[i][j]=='*')
{
if(c[i]+d[j]==sum+1)
{
sx=i+1;
sy=j+1;
flag=1;
}
}
else
{
if(c[i]+d[j]==sum)
{
sx=i+1;
sy=j+1;
flag=1;
}
}
if(flag==1)
{
break;
}
}
if(flag==1)
{
break;
}
}
if(flag==1)
{
printf("YES\n");
printf("%d %d\n",sx,sy);
}
else
{
printf("NO\n");
}
}
return 0;
}