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>
const long long mod = 1e9 + 7;
const int sz = 100010;
int a[sz];
struct ele {
  int l, x, id;
} c[sz];
int ans[sz];
long long fast(long long x, long long a) {
  if (x == 0) return 1;
  long long y = fast(x / 2, a);
  y *= y;
  y %= mod;
  if (x & 1) return (a * y) % mod;
  return y;
}
bool cmp(ele p, ele q) { return p.l < q.l; }
int ins[(1 << 20)], S[(1 << 20)];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int k = 1; k <= n; k++) scanf("%d", &a[k]);
  for (int k = 1; k <= q; k++) {
    scanf("%d%d", &c[k].l, &c[k].x);
    c[k].id = k;
  }
  std::sort(c + 1, c + q + 1, cmp);
  int cnt = 0, in = 0;
  S[0] = 0;
  ins[0] = 1;
  for (int k = 1; k <= q; k++) {
    while (in < c[k].l) {
      in++;
      if (ins[a[in]]) continue;
      for (int i = 0; i < (1 << cnt); i++) {
        int x = S[i] ^ a[in];
        ins[x] = 1;
        S[(1 << cnt) + i] = x;
      }
      cnt++;
    }
    if (ins[c[k].x]) {
      ans[c[k].id] = fast((long long)c[k].l - cnt, 2);
    } else
      ans[c[k].id] = 0;
  }
  for (int k = 1; k <= q; k++) printf("%d\n", ans[k]);
}
