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;
int n, qr;
vector<pair<int, int> > q[100005];
vector<int> Set;
int l, x, v[100005], sol[100005];
bool inSet[10000000];
int main() {
  cin >> n >> qr;
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1; i <= qr; i++) {
    scanf("%d%d", &l, &x);
    q[l].push_back({x, i});
  }
  Set.push_back(0);
  inSet[0] = 1;
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    if (inSet[v[i]])
      ans = (ans * 2) % 1000000007;
    else {
      int sz = Set.size();
      for (int j = 0; j < sz; j++) {
        Set.push_back(Set[j] ^ v[i]);
        inSet[Set.back()] = 1;
      }
    }
    for (auto it : q[i]) {
      sol[it.second] = ans * inSet[it.first];
    }
  }
  for (int i = 1; i <= qr; i++) printf("%d\n", sol[i]);
  return 0;
}
