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;
vector<int> v;
int a[200000];
struct edge {
  int l;
  int x;
  int mark;
  long long ans;
} e[200000];
int vis[10000000];
int cmp(edge a, edge b) { return a.l < b.l; }
int cmp2(edge a, edge b) { return a.mark < b.mark; }
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &e[i].l, &e[i].x);
    e[i].mark = i;
  }
  sort(e, e + q, cmp);
  v.push_back(0);
  int now = 0;
  vis[0] = 1;
  int temp = 1;
  for (int i = 0; i < q; i++) {
    while (now < e[i].l) {
      if (vis[a[now]]) {
        temp = (temp * 2) % 1000000007;
      } else {
        int x = v.size();
        for (int ii = 0; ii < x; ii++) {
          v.push_back(v[ii] ^ a[now]);
          vis[v[ii] ^ a[now]] = 1;
        }
      }
      now++;
    }
    e[i].ans = vis[e[i].x] * temp;
  }
  sort(e, e + q, cmp2);
  for (int i = 0; i < q; i++) cout << e[i].ans << endl;
  return 0;
}
