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 mod = 1000000007;
bool f[1 << 20 + 5];
struct Query {
  int l, x, ind, sol;
} q[100005];
int v[100005];
bool cmp(Query a, Query b) { return a.l < b.l; }
bool cmp1(Query a, Query b) { return a.ind < b.ind; }
int main() {
  int n, cer, i, frecv, j, val;
  scanf("%d%d", &n, &cer);
  for (i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (i = 1; i <= cer; i++) {
    scanf("%d%d", &q[i].l, &q[i].x);
    q[i].ind = i;
  }
  sort(q + 1, q + cer + 1, cmp);
  frecv = 1;
  j = 1;
  f[0] = true;
  for (i = 1; i <= n; i++) {
    if (!f[v[i]]) {
      f[v[i]] = true;
      for (val = 0; val < (1 << 20); val++)
        if (f[val]) f[val ^ v[i]] = true;
    } else
      frecv = (frecv * 1LL * 2) % mod;
    while (j <= cer && q[j].l == i) {
      if (f[q[j].x])
        q[j].sol = frecv;
      else
        q[j].sol = 0;
      j++;
    }
  }
  sort(q + 1, q + cer + 1, cmp1);
  for (i = 1; i <= cer; i++) printf("%d\n", q[i].sol);
  return 0;
}
