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 maxn = 100000 + 100;
const int MOD = 1000000000 + 7;
int n, q, s[maxn][20], c[maxn], pow2[maxn];
void work() {
  scanf("%d%d", &n, &q);
  pow2[0] = 1;
  for (int i = 1, x; i <= n; i++) {
    c[i] = c[i - 1];
    pow2[i] = pow2[i - 1] * 2 % MOD;
    memcpy(s[i], s[i - 1], sizeof(s[i]));
    scanf("%d", &x);
    for (int j = 0; j < 20; j++) {
      if (((x >> j) & 1) && (s[i][j] > 0)) {
        x ^= s[i][j];
      }
    }
    if (x == 0) continue;
    for (int j = 0; j < 20; j++) {
      if ((x >> j) & 1) {
        s[i][j] = x;
        c[i]++;
        break;
      }
    }
  }
  while (q--) {
    int l, x;
    scanf("%d%d", &l, &x);
    for (int i = 0; i < 20; i++) {
      if ((x >> i) & 1) {
        x ^= s[l][i];
      }
    }
    if (x != 0) {
      printf("0\n");
    } else {
      printf("%d\n", pow2[l - c[l]]);
    }
  }
}
int main() {
  int TestCase = 1;
  while (TestCase--) {
    work();
  }
  return 0;
}
