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 inf = 0x7fffffff;
int c[100009];
int a[100009];
int cnt = 0;
void ins(int x) {
  for (int i = 21; i >= 0; i--) {
    if (x & (1 << i)) {
      if (c[i]) {
        x ^= c[i];
      } else {
        c[i] = x;
        cnt++;
        return;
      }
    }
  }
}
int n, q;
struct node {
  int l, x, id;
} b[100009];
int cf[100009];
int cmp(node a, node b) {
  if (a.l != b.l) return a.l < b.l;
  return a.x < b.x;
}
bool check(int x) {
  for (int i = 21; i >= 0; i--) {
    if (x & (1 << i)) {
      if (c[i] == 0) return 0;
      x ^= c[i];
    }
  }
  return 1;
}
int ans[100009];
signed main() {
  scanf("%d%d", &n, &q);
  cf[0] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    cf[i] = (cf[i - 1] * 2) % 1000000007;
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &b[i].l, &b[i].x);
    b[i].id = i;
  }
  sort(b + 1, b + 1 + q, cmp);
  int p = 0;
  for (int i = 1; i <= q; i++) {
    while (p < b[i].l) {
      ins(a[++p]);
    }
    if (check(b[i].x)) {
      ans[b[i].id] = cf[p - cnt];
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
