对于每一组测试数据，输出需要的最少步数。

#include<iostream>

#include<cstring>

#include<cstdio>

const int N =  + ;

char mat[N];

int main(){

    int T;

    scanf("%d", &T);

    while(T --){

        int n, m;

        scanf("%d %d", &n, &m);

        for(int i = ; i < m; i++) scanf("%s", mat);

        if(n ==  && m == ) puts("");

        else if(n == ) printf("%d\n",  * (m - ));

        else if(m == ) printf("%d\n",  * (n - ));

        else if(n %  ==  || m %  == ) printf("%d\n", n * m);

        else printf("%d\n", n * m + );

    }

    return ;

}