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 long long mod = 1e9 + 7;
int n, q;
int a[100001];
long long p2[100001];
vector<pair<int, int> > qr[100001];
long long ans[100001];
long long boss[20];
int cnt = 0;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q;
  p2[0] = 1;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    p2[i] = p2[i - 1] * 2 % mod;
  }
  for (int i = 1; i <= q; i++) {
    int u, v;
    cin >> u >> v;
    qr[u].push_back(make_pair(v, i));
  }
  for (int i = 1; i <= n; i++) {
    int cur = a[i];
    for (int j = 19; j >= 0; j--) {
      if (a[i] & (1 << j)) {
        if (boss[j] == 0) {
          boss[j] = a[i];
          break;
        } else
          a[i] ^= boss[j];
      }
    }
    if (a[i] == 0) cnt++;
    for (int j = 0; j < qr[i].size(); j++) {
      bool hv = true;
      for (int k = 19; k >= 0; k--) {
        if (qr[i][j].first & (1 << k)) {
          if (boss[k] == 0) {
            hv = false;
            break;
          }
          qr[i][j].first ^= boss[k];
        }
      }
      if (hv) ans[qr[i][j].second] = p2[cnt];
    }
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
}
