題意：

問有多少個n*m的012矩陣，存在至少一組 ( x 1 , x 2 ) ( y 1 , y 1 ) (x1,x2)(y1,y1) (x1,x2)(y1,y1)使得4個點 ( x 1 , y 1 ) ( x 1 , y 2 ) ( x 2 , y 1 ) ( x 2 , y 2 ) (x1,y1)(x1,y2)(x2,y1)(x2,y2) (x1,y1)(x1,y2)(x2,y1)(x2,y2)組成
（0和1可以用2代替，也就是兩列各自相同或者兩行各自相同）

解析：

分析一下n*2：

∣ 0 , 0 ∣ |0,0| ∣0,0∣
∣ 0 , 1 ∣ |0,1| ∣0,1∣
∣ 0 , 2 ∣ |0,2| ∣0,

得出當n或者m大於7時一定存在，所以我們只需要暴力求出n<8,m<8的情況即可

程式碼：

/*
 *  Author : Jk_Chen
 *    Date : 2020-12-25-12.46.27
 */
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define LD long double
#define rep(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
 

#define per(i,a,b) for(int i=(int)(a);i>=(int)(b);i--)
#define mmm(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define pill pair<int, int>
#define fi first
#define se second
void test(){cerr<<"\n";}
template<typename T,typename... Args>void test(T x,Args... args){
 
cerr<<"> "<<x<<" ";test(args...);}
const LL _mod=1e9+7;
const int maxn=1e5+9;
const int inf=0x3f3f3f3f;
LL rd(){ LL ans=0; char last=' ',ch=getchar();
    while(!(ch>='0' && ch<='9'))last=ch,ch=getchar();
    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
    if(last=='-')ans=-ans; return ans;
}
#define rd rd()
/*_________________________________________________________begin*/

LL Pow(LL a,LL b,LL mod=_mod){
    if(a>=mod)a%=mod;
    LL res=1;
    while(b>0){
        if(b&1)res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
/*_________________________________________________________Pow*/

LL ans[10][10];
int mp[10][10];
int n,m;

void dfs(int i,int j){
    if(i==n+1){
        ans[n][m]++;
        return;
    }
    rep(k,0,2){
        bool can=1;
        rep(_i,1,i-1)
            rep(_j,1,j-1){
                if(mp[_i][_j]==mp[_i][j]&&mp[i][_j]==k){
                    _i=i;can=0;break;
                }
                if(mp[_i][_j]==mp[i][_j]&&mp[_i][j]==k){
                    _i=i;can=0;break;
                }
            }
        if(can){
            mp[i][j]=k;

            int toi=i;
            int toj=j;

            if(j==m){
                toi++;
                toj=1;
            }
            else{
                toj++;
            }
            dfs(toi,toj);
        }
    }
}

void init(int _n,int _m){
    n=_n,m=_m;
    dfs(1,1);
}

LL A[8][8]={
0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,
0,0,66,390,1800,6120,13680,15120,
0,0,390,3198,13176,27000,13680,15120,
0,0,1800,13176,24336,4320,0,0,
0,0,6120,27000,4320,4320,0,0,
0,0,13680,13680,0,0,0,0,
0,0,15120,15120,0,0,0,0,
};

int main(){
//    rep(i,2,7)rep(j,2,7){
//        init(i,j);
//        //printf("%d,%d) %lld\n",i,j,ans[i][j]);
//    }
//    rep(i,1,7)rep(j,1,7){
//        printf("%lld%c",ans[i][j],",\n"[j==7]);
//    }
    int t=rd;
    while(t--){
        n=rd,m=rd;
        if(n==1||m==1){
            printf("%lld\n",0ll);
        }
        else if(n>7||m>7){
            printf("%lld\n",Pow(3,n*m));
        }
        else{
            printf("%lld\n",(Pow(3,n*m)+_mod-A[n][m])%_mod);
        }
    }
    return 0;
}

/*_________________________________________________________end*/