#include <bits/stdc++.h>
//#include <atcoder/modint>

using namespace std;
//using namespace atcoder;
using ll = long long;
//using mint = modint998244353;

struct UnionFind {
    int ngroup, N;
    vector<int> par, siz;

    UnionFind(int _N) : N(_N), par(_N), siz(_N){
        ngroup = _N;
        for(int i = 0; i < N; i++) par[i] = i, siz[i] = 1;
    }

    inline int root(int x) {
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    int unite(int x, int y) {
        int rx = root(x), ry = root(y);
        if (rx == ry) return rx;
        ngroup--;
        if (siz[rx] > siz[ry]) swap(rx, ry);
        par[rx] = ry; siz[ry] += siz[rx];
        return ry;
    }

    inline bool same(int x, int y) {
        int rx = root(x), ry = root(y);
        return rx == ry;
    }

    inline int size(int x) {return siz[root(x)];}
    inline int group_count() const{return ngroup;}

    vector<vector<int>> groups() {
        vector<int> rev(N); int nrt=0;
        for (int i=0; i<N; i++) if (root(i) == i) rev[i] = nrt, nrt++;
        vector<vector<int>> res(nrt);
        for (int i=0; i<N; i++) res[rev[root(i)]].push_back(i);

        return res;
    }
};


int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    /*
       各マスにクエリを持たせておいて、マスの小さい順にマージしていく。
       連結成分でクエリの2つのペアが揃うなら、その時点でのマスの整数の最大値とする。
       マージテクでO(QlogHW+HW)
    */

    int H, W, N, Q, x, y, z, w;
    cin >> H >> W;
    N = H*W;
    UnionFind tree(N);
    vector a(H, vector<int>(W));
    vector<tuple<int, int, int>> v;
    for (int i=0; i<H; i++){
        for (int j=0; j<W; j++){
            cin >> a[i][j];
            v.push_back({a[i][j], i, j});
        }
    }
    sort(v.begin(), v.end());
    cin >> Q;
    vector<unordered_set<int>> qr(N);
    vector<int> s(Q), t(Q), ans(Q);
    for (int i=0; i<Q; i++){
        cin >> x >> y >> z >> w;
        x--; y--; z--; w--;
        x = x*W+y;
        z = z*W+w;
        s[i] = x;
        t[i] = z;
        qr[x].insert(i);
        qr[z].insert(i);
    }

    auto merge=[&](int p, int q, int now)->void{
        p = tree.root(p); q = tree.root(q);
        if (p == q) return;
        if (qr[p].size() > qr[q].size()) swap(p, q);
        for (auto i : qr[p]){
            if (qr[q].count(i)){
                qr[q].erase(i);
                ans[i] = now;
            }
            else qr[q].insert(i);
        }
        qr[p].clear();
        int r = tree.unite(p, q);
        if (q != r) swap(qr[q], qr[r]);
    };

    int dx[4] = {1,0,-1,0};
    int dy[4] = {0,1,0,-1};
    auto check=[&](int i, int j){
        return 0 <= i && i < H && 0 <= j && j < W;
    };
    for (auto [c, i, j] : v){
        for (int k=0; k<4; k++){
            int ni = i+dx[k], nj = j+dy[k];
            if (check(ni, nj) && a[ni][nj] <= a[i][j]){
                merge(i*W+j, ni*W+nj, c);
            }
        }
    }
    for (int i=0; i<Q; i++) cout << ans[i] << '\n';

    return 0;
}