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 long long MOD = 1000000007;
const int MAX_N = 100010;
const int d = 35;
int n, q;
int arr[MAX_N];
int basisAtT[MAX_N][d];
int basis[d];
long long p2[MAX_N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  p2[0] = 1;
  for (int i = 1; i < MAX_N; i++) p2[i] = (p2[i - 1] * 2LL) % MOD;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) {
    int val = arr[i];
    while (val != 0) {
      int idx = __builtin_clz(val);
      if (basis[idx] != 0) {
        val ^= basis[idx];
      } else {
        ;
        basis[idx] = val;
        break;
      }
    }
    for (int j = 0; j < d; j++) basisAtT[i][j] = basis[j];
  }
  for (int i = 0; i < d; i++)
    ;
  while (q--) {
    int l, x;
    cin >> l >> x;
    l--;
    bool covered = 1;
    int val = x;
    while (val != 0) {
      int idx = __builtin_clz(val);
      if (basisAtT[l][idx] != 0) {
        val ^= basisAtT[l][idx];
      } else {
        covered = 0;
        break;
      }
    }
    if (covered) {
      int bsize = 0;
      for (int i = 0; i < d; i++) bsize += (basisAtT[l][i] != 0);
      int cnt = l + 1 - bsize;
      cout << p2[cnt] << "\n";
    } else {
      cout << "0\n";
    }
  }
}
