codeforces 837B Flag of Berland

B. Flag of Berland

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The flag of Berland is such rectangular field n × m that satisfies following conditions:

• Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
• Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
• Each color should be used in exactly one stripe.

You are given a field n × m, consisting of characters 'R', 'G' and 'B'. Output "YES" (without quotes) if this field corresponds to correct flag of Berland. Otherwise, print "NO" (without quotes).

Input

The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.

Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field.

Output

Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes).

Examples

input

6 5RRRRRRRRRRBBBBBBBBBBGGGGGGGGGG

output

YES

input

4 3BRGBRGBRGBRG

output

YES

input

6 7RRRGGGGRRRGGGGRRRGGGGRRRBBBBRRRBBBBRRRBBBB

output

NO

input

4 4RRRRRRRRBBBBGGGG

output

NO

Note

The field in the third example doesn't have three parralel stripes.

Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1.

感觉这类题目应该配个图，单纯看英文不怎么好理解。题意就是给出一幅图，包含3中颜色，由3个大写字母表示。问该图是否符合类似于俄罗斯国旗一样的3色分布。

并且认为竖着的也可以。

可以先默认的横着的去check一下，check可以先去判断行数能否整除三然后判断是否是3种颜色都是在行数上连续的，然后看每一行是否都等于该行首。

然后就是把图翻转90度再check，check仍然如上所述，最终得到结果。

#include<bits/stdc++.h>using namespace std;char ma[111][111];char ma1[111][111];int n,m;bool ch(int x,int y){if(x%3!=0)return 0;int d=x/3;if(ma[0][0]==ma[d][0]||ma[0][0]==ma[d*2][0]||ma[d][0]==ma[d*2][0])return 0;for(int i=0;i<n;i++)for(int j=0;j<m;j++)if(ma[i][j]!=ma[i/d*d][0])return 0;return 1;}int main(){   while(cin>>n>>m)   {   int i,j;   for(i=0;i<n;i++)   cin>>ma[i];   if(ch(n,m))    {    cout<<"YES"<<endl;}else {for(i=0;i<n;i++)for(j=0;j<m;j++)ma1[j][i]=ma[i][j];swap(n,m);for(i=0;i<n;i++)for(j=0;j<m;j++)ma[i][j]=ma1[i][j];if(ch(n,m)){cout<<"YES"<<endl;}else cout<<"NO"<<endl;}   }return 0;}