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<pair<int, int> > v[100005];
int arr[100005], a[100005];
vector<int> s;
bool b[(1 << 20)];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  for (int i = 0; i < q; i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    v[l - 1].push_back({x, i});
  }
  s.push_back(0);
  b[0] = 1;
  int ans = 1;
  for (int i = 0; i < n; i++) {
    if (b[arr[i]])
      ans = (ans * 2) % 1000000007;
    else {
      int tmp = s.size();
      for (int x = 0; x < tmp; x++) {
        s.push_back(s[x] ^ arr[i]);
        b[s.back()] = 1;
      }
    }
    for (int x = 0; x < v[i].size(); x++)
      a[v[i][x].second] = ans * b[v[i][x].first];
  }
  for (int i = 0; i < q; i++) printf("%d\n", a[i]);
}
