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 = 1e9 + 7;
int fi[100005], ans[100005], a[100005];
vector<pair<int, int> > query[100005];
int main() {
  memset(fi, -1, sizeof(fi));
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= q; i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    query[l].push_back(pair<int, int>(i, x));
  }
  int res = 1;
  for (int i = 1; i <= n; i++) {
    bool ok = 0;
    for (int j = 20; j >= 0; j--) {
      if ((a[i] >> j) & 1) {
        if (fi[j] == -1) {
          fi[j] = a[i];
          ok = 1;
          break;
        } else
          a[i] ^= fi[j];
      }
    }
    if (!ok) res = (res * 2) % mod;
    for (int j = 0; j < query[i].size(); j++) {
      int idx = query[i][j].first, x = query[i][j].second;
      bool ok = 1;
      for (int j = 20; j >= 0; j--) {
        if ((x >> j) & 1) {
          if (fi[j] == -1) {
            ok = 0;
            break;
          }
          x ^= fi[j];
        }
      }
      if (ok) ans[idx] = res;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
