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 N = 100005;
const long long mod = 1000000007;
int basis[N][20], basis_sz[N];
long long pw2[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  int n, q, l, x;
  cin >> n >> q;
  memset(basis, -1, sizeof(basis));
  basis_sz[0] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    for (int j = 0; j < 20; j++) {
      basis[i][j] = basis[i - 1][j];
      if (basis[i][j] != -1) {
        if (x & (1 << j)) {
          x ^= basis[i][j];
        }
      }
    }
    basis_sz[i] = basis_sz[i - 1];
    for (int j = 0; j < 20; j++) {
      if (x & (1 << j)) {
        assert(basis[i][j] == -1);
        basis[i][j] = x;
        basis_sz[i]++;
        break;
      }
    }
  }
  pw2[0] = 1;
  for (int i = 1; i < N; i++) {
    pw2[i] = (pw2[i - 1] << 1);
    if (pw2[i] >= mod) pw2[i] -= mod;
  }
  while (q--) {
    cin >> l >> x;
    for (int i = 0; i < 20; i++) {
      if (x & (1 << i)) {
        if (basis[l][i] != -1) x ^= basis[l][i];
      }
    }
    if (x == 0) {
      cout << pw2[l - basis_sz[l]] << '\n';
    } else {
      cout << 0 << '\n';
    }
  }
  return 0;
}
