由于 $f(l, r) = a_l \And a_{l + 1} \ldots \And a_{r}$ 是有单调性的，因此可以考虑二分答案。接下来是如何快速求 $f(l, r)$，可以按位考虑，对每一个二进制位，计算区间内这个二进制位为 1 的个数，如果恰好等于区间的长度，说明区间内所有数字的这位都为 1，累加到答案中。数区间内 1 的个数可以使用前缀和快速得到。

// Date: Wed Dec 27 23:50:18 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
  for (t &x : vec)
    in >> x;
  return in;
}

template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
  int n = SZ(vec);
  For(i, 0, n) {
    out << vec[i];
    if (i < n - 1)
      out << ' ';
  }
  return out;
}

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 200010;
int t, q, a[N], d[N][35], l, k, n;

int check(int l, int r) {
  int res = 0;
  For(j, 0, 32) {
    int cn = d[r][j] - d[l - 1][j];
    if (cn == r - l + 1) {
      res += (1 << j);
    }
  }
  return res;
}

int main(void) {
#ifdef _DEBUG
  freopen("1878e.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;
  while (t--) {
    cin >> n;

    For1(i, 1, n) { cin >> a[i]; }
    cin >> q;

    For(j, 0, 32) d[0][j] = 0;

    For1(i, 1, n) {
      int tmp = a[i];
      For(j, 0, 32) {
        int cnt = 0;
        if (tmp & 1)
          cnt = 1;
        tmp >>= 1;

        d[i][j] = d[i - 1][j] + cnt;
      }
    }

    while (q--) {
      cin >> l >> k;
      int r = n, mid;

      int le = l;
      while (le < r) {
        mid = (le + r + 1) / 2;
        if (check(l, mid) >= k)
          le = mid;
        else
          r = mid - 1;
      }
      if (check(l, le) >= k)
        cout << le;
      else
        cout << -1;
      cout << ' ';
    }
    cout << '\n';
  }

  return 0;
}