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;
int sz[N], p[N][30], pot[N];
bool brute(int i, int x) {
  for (int j = 0, b = 19; b >= 0; b--) {
    if (!(p[i][j] & (1 << b))) {
      if (x & (1 << b)) return 0;
      continue;
    }
    if (x & (1 << b)) x ^= p[i][j];
    j++;
  }
  return !x;
}
int main() {
  pot[0] = 1;
  for (int i = 1; i < N; i++) {
    pot[i] = pot[i - 1] << 1;
    if (pot[i] >= 1000000007) pot[i] -= 1000000007;
  }
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    memcpy(p[i], p[i - 1], sizeof p[i]), sz[i] = sz[i - 1];
    int b, j;
    for (b = 19, j = 0; b >= 0 && j < sz[i]; b--) {
      int k = j;
      while (k < sz[i] && !(p[i][k] & (1 << b))) k++;
      if (sz[i] == k) continue;
      swap(p[i][j], p[i][k]);
      if (x & (1 << b)) x ^= p[i][j];
      j++;
    }
    for (b = 19; b >= 0; b--)
      if (x & (1 << b)) {
        for (j = 0; j < sz[i]; j++)
          if (p[i][j] & (1 << b)) p[i][j] ^= x;
        break;
      }
    j = 0;
    for (int k = 0; k < sz[i]; k++)
      if (p[i][k]) p[i][j++] = p[i][k];
    sz[i] = j;
    if (x) p[i][sz[i]++] = x;
    sort(p[i], p[i] + sz[i]);
    reverse(p[i], p[i] + sz[i]);
  }
  while (q--) {
    int l, x;
    scanf("%d %d", &l, &x);
    printf("%d\n", brute(l, x) ? pot[l - sz[l]] : 0);
  }
  return 0;
}
