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 N = (int)1e5 + 7;
const int inf = (int)1e9 + 7;
bool b[1 << 20];
vector<pair<int, int> > query[N];
int ans[N];
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  long long res = 1;
  vector<int> v;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    v.push_back(a);
  }
  for (int i = 1; i <= q; i++) {
    int l, x;
    scanf("%d %d", &l, &x);
    query[l - 1].push_back(make_pair(x, i));
  }
  b[0] = 1;
  vector<int> s;
  s.push_back(0);
  for (int i = 0; i < n; i++) {
    if (b[v[i]]) {
      res <<= 1;
      res %= inf;
    } else {
      int siz = (int)s.size();
      for (int j = 0; j < siz; j++) {
        if (!b[s[j] ^ v[i]]) {
          b[s[j] ^ v[i]] = 1;
          s.push_back(s[j] ^ v[i]);
        }
      }
    }
    for (int j = 0; j < (int)query[i].size(); j++) {
      ans[query[i][j].second] = (b[query[i][j].first] ? res : 0);
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
}
