Ehab has an array a of n integers. He likes the [bitwise-xor operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) and he likes to bother Mahmoud so he came up with a problem. He gave Mahmoud q queries. In each of them, he gave Mahmoud 2 integers l and x, and asked him to find the number of subsequences of the first l elements of the array such that their bitwise-xor sum is x. Can you help Mahmoud answer the queries?

A subsequence can contain elements that are not neighboring.

Input

The first line contains integers n and q (1 ≤ n, q ≤ 105), the number of elements in the array and the number of queries.

The next line contains n integers a1, a2, ..., an (0 ≤ ai < 220), the elements of the array.

The next q lines, each contains integers l and x (1 ≤ l ≤ n, 0 ≤ x < 220), representing the queries.

Output

For each query, output its answer modulo 109 + 7 in a newline.

Examples

Input

5 5
0 1 2 3 4
4 3
2 0
3 7
5 7
5 8


Output

4
2
0
4
0


Input

3 2
1 1 1
3 1
2 0


Output

4
2

Note

The bitwise-xor sum of the empty set is 0 and the bitwise-xor sum of a set containing one element is that element itself.
#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
struct xor_basis {
  int d, sz;
  vector<int> basis;
  xor_basis(int d) {
    this->d = d;
    sz = 0;
    basis.resize(d);
  }
  bool add(int mask, int actually_add) {
    if (mask == 0) return 0;
    for (int i = 0; i < d; i++) {
      if (mask & (1 << i)) {
        if (!basis[i]) {
          if (actually_add) {
            basis[i] = mask;
            sz++;
          }
          return 1;
        }
        mask ^= basis[i];
      }
    }
    return 0;
  }
};
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  int rs = 1;
  vector<tuple<int, int, int>> evs;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    evs.emplace_back(i, 0, x);
  }
  for (int i = 0; i < q; i++) {
    int l, x;
    cin >> l >> x;
    evs.emplace_back(l - 1, i + 1, x);
  }
  sort(evs.begin(), evs.end());
  vector<int> ans(q);
  xor_basis xrb(20);
  for (auto [l, id, x] : evs) {
    if (id == 0) {
      if (!xrb.add(x, 1)) {
        rs = (rs * 2) % M;
      }
    } else {
      id--;
      ans[id] = rs * (!xrb.add(x, 0));
    }
  }
  for (int i : ans) cout << i << '\n';
  return 0;
}
