CodeForces - 404A: Valera and X

A. Valera and X

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals n squares (n is an odd number) and each unit square contains some small letter of the English alphabet.

Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:

• on both diagonals of the square paper all letters are the same;
• all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.

Help Valera, write the program that completes the described task for him.

Input

The first line contains integer n (3 ≤ n < 300; n is odd). Each of the next n lines contains n small English letters — the description of Valera's paper.

Output

Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.

Examples

input

5xoooxoxoxosoxoooxoxoxooox

output

NO

input

3wswswswsw

output

YES

input

3xpxpxpxpe

output

NO

大致题意是对角线要同一种字符，其余位置是另外一种字符。

# include <stdio.h># include <string.h>int main(){    int n, i, j ,flag = 0;    char s[302][302], c, d;    scanf("%d",&n);    for(i=1; i<=n; ++i)        scanf("%s",s[i]+1);    c = s[1][1];//保存第一个字符    d = s[1][2];//保存第二个字符    if(c == d)    {        puts("NO");        return 0;    }    for(i=1; i<=n/2; ++i)//先遍历到中间行    {        if(s[i][i] != s[i][n-i+1] || s[i][i] != c || s[i][i]==d)            flag = 1;        for(j=1; j<=n; ++j)        {            if(j==i || j==n-i+1)                continue;            else if(s[i][j] != d)            {                flag = 1;                break;            }        }        if(flag)            break;    }    for(i=n/2+1; i<=n; ++i)//再遍历后面的行    {        if(s[i][n-i+1] != s[i][i] || s[i][i] != c || s[i][i]==d)            flag = 1;        for(j=1; j<=n; ++j)        {            if(j==n-i+1 || j==i)                continue;            else if(s[i][j] != d)            {                flag = 1;                break;            }        }        if(flag)            break;    }    if(flag)        puts("NO");    else        puts("YES");    return 0;}