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 n, q;
int p[100005], a[100005], bas[100005][20], cnt[100005];
void construct(int id) {
  cnt[id] = cnt[id - 1];
  for (int i = 0; i < 20; i++) bas[id][i] = bas[id - 1][i];
  int x = a[id];
  for (int j = 19; j >= 0; j--) {
    if (x & (1 << j)) {
      if (!bas[id][j]) {
        bas[id][j] = x;
        cnt[id]++;
        break;
      }
      x ^= bas[id][j];
    }
  }
}
int query(int id, int x) {
  for (int j = 19; j >= 0; j--) {
    if (x & (1 << j)) {
      if (!bas[id][j]) return 0;
      x ^= bas[id][j];
    }
  }
  return p[id - cnt[id]];
}
int main() {
  p[0] = 1;
  for (int i = 1; i <= 100000; i++) p[i] = 2LL * p[i - 1] % 1000000007;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    construct(i);
  }
  for (int i = 0; i < q; i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    printf("%d\n", query(l, x));
  }
  return 0;
}
