#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#define ll long long
#define rep(i,n) for(ll i=0;i<(n);i++)
#define pll pair<ll,ll>
#define pq priority_queue
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define lb(c,x) distance(c.begin(),lower_bound(all(c),x))
#define ub(c,x) distance(c.begin(),upper_bound(all(c),x))

using namespace std;

template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}

const ll INF=1e18;
const ll mod=1e9+7;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(ll n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (ll i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (ll i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(ll n, ll k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
  mint p(ll n, ll k) {
    return fact[n]*ifact[n-k];
  }
} c(1000005);

int main(){
    ll h,w;
    cin >> h >> w;
    vector<vector<ll>> a(h,vector<ll>(w));
    vector<ll> R(h),C(w),mul_r(h,1),mul_c(w,1);
    ll mul=1;
    rep(i,h){
        rep(j,w){
            cin >> a[i][j];
        }
    }
    ll tot=0;
    rep(i,h){
        rep(j,w){
            if(a[i][j]==0){
                R[i]++;
                C[j]++;
                tot++;
            }
            mul*=max(a[i][j],1LL);
            mul%=mod;
        }
    }
    rep(i,h){
        rep(j,w){
            mul_r[i]*=max(a[i][j],1LL);
            mul_r[i]%=mod;
        }
    }
    rep(i,w){
        rep(j,h){
            mul_c[i]*=max(a[j][i],1LL);
            mul_c[i]%=mod;
        }
    }
    ll q;
    cin >> q;
    rep(i,q){
        ll r,c;
        cin >> r >> c;
        r--; c--;
        ll now_zero=0;
        if(a[r][c]==0) now_zero=1;
        if(tot>R[r]+C[c]-now_zero){
            cout << 0 << endl;
        }
        else{
            mint rr=mul_r[r];
            mint cc=mul_c[c];
            mint buf=rr*cc;
            mint num=max(a[r][c],1LL);
            buf/=num;
            mint ans=mul;
            ans/=buf;
            cout << ans.x << endl;
        }
    }
    return 0;
}