Let $g(i)$ be $f(L, i)$, consider each bit separately:
If this bit is $0$, then after every subsequent $and$ it remains $0$.
If this bit is $1$, either it remains 1 after every subsequent $and$, or there exists a certain $and$ after which the result is $0$, and it remains $0$ thereafter. Thus, $g$ is a monotonically decreasing function. Binary search, $O(m\log^2)$.

There is another method, preprocess and find out for each bit the position where it changes from $1$ to $0$. For each query, start looping from the highest bit, time complexity $O(n\log + m\log)$

// compile: make data
// run: ./data < data.in
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#ifdef LOCAL
#include <debug/codeforces.h>
#define debug(x...) _debug_print(#x, x);
#define Debug(x...) _debug_print_format(#x, x);
std::ifstream terminal("/dev/tty");
#define PP cerr<<"\033[1;30mpause...\e[0m",terminal.ignore();
#else
#define debug(x...)
#define Debug(x...)
#define PP
#endif
template<typename...Args> void print_(Args...args){((cout<<args<<" "),...)<<endl;}
#define rep(i,a,b) for(int i=(a);i<(int)(b);++i)
#define sz(v) ((int)(v).size())
#define print(...) print_(__VA_ARGS__);
#define FIND(a, x) ((find(a.begin(),a.end(),(x))!=a.end())?1:0)
#define cmin(x,...) x=min({(x),__VA_ARGS__})
#define cmax(x,...) x=max({(x),__VA_ARGS__})
#define INTMAX (int)(9223372036854775807)
#define INF (int)(1152921504606846976)
#define NaN (int)(0x8b88e1d0595d51d1)
#define double long double
#define int long long
#define uint unsigned long long
#define MAXN 200010

int tonum(vector<int> &v, int off) {
    int ans = 0;
    for (int i = 0; i < sz(v); i++) {
        if (v[i] >= off) ans += (1 << i);
    }
    return ans;
}

int bs(vector<vector<int>> &dp, int l, int r, int k) {
    int ans = -1;
    int L = l;
    while (l <= r) {
        vector<int> offset(32, 0);
        int mid = (l + r) / 2;
        rep(i, 0, 32) offset[i] = dp[mid][i] - dp[L-1][i];
        int num = tonum(offset, mid-L+1);
        // debug(l, mid, r, num)
        if (num >= k) ans = mid, l = mid + 1;
        else r = mid - 1;
    }
    return ans;
}

void solve() {
    int n; cin >> n;
    vector<bitset<32>> a(n+1);
    rep(i, 1, n+1) {
        int t; cin >> t;
        a[i] = t;
    }
    vector<vector<int>> dp(n+1, vector<int>(32, 0));
    rep(i, 1, n+1) {
        rep(j, 0, 32) {
            dp[i][j] = dp[i-1][j] + a[i][j];
        }
    }
    int q; cin >> q;
    // vector<int> offset(32, 0);
    // rep(i, 0, 32) offset[i] = dp[3][i] - dp[0][i];
    // int num = tonum(offset, 3);
    // debug(num)
    while (q--) {
        int l, k; cin >> l >> k;
        int idx = bs(dp, l, n, k);
        cout << idx << " ";
        // break;
    }
    cout << endl;
}

int32_t main() {
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);

    int _; cin >> _;
    while (_--) solve();

    return 0;
}