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;
int p[100005][21], sz[100005], n, q;
int pw[100005];
int main() {
  int i, j, l, x;
  scanf("%d%d", &n, &q);
  pw[0] = 1;
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    sz[i] = sz[i - 1];
    pw[i] = pw[i - 1] + pw[i - 1];
    if (pw[i] >= 1000000007) pw[i] -= 1000000007;
    memcpy(p[i], p[i - 1], sizeof(p[i - 1]));
    for (j = 19; j >= 0; j--) {
      if (x & (1 << j)) {
        if (!p[i][j]) {
          p[i][j] = x;
          sz[i]++;
          break;
        } else {
          x ^= p[i][j];
        }
      }
    }
  }
  while (q--) {
    scanf("%d%d", &l, &x);
    for (i = 19; i >= 0; i--) {
      if (x & (1 << i)) x ^= p[l][i];
    }
    if (x) {
      puts("0");
      continue;
    }
    printf("%d\n", pw[l - sz[l]]);
  }
  return 0;
}
