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>
const int MX = 1 << 21, MD = 1e9 + 7;
using namespace std;
int n, q, ans[MX], a[MX], l[MX], X[MX];
vector<int> qw[MX];
int main() {
  cin >> n >> q;
  vector<int> V(1, 0);
  ans[0] = 1;
  bitset<1 << 21> B;
  B[0] = true;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= q; i++) {
    cin >> l[i] >> X[i];
    qw[l[i]].push_back(X[i]);
  }
  map<pair<int, int>, int> Nig;
  for (int i = 1; i <= n; i++) {
    if (B[a[i]]) {
      ans[i] = (ans[i - 1] + ans[i - 1]) % MD;
    } else {
      int oldSize = V.size();
      for (int j = 0; j < oldSize; j++) {
        V.push_back(a[i] ^ V[j]);
        B[a[i] ^ V[j]] = 1;
      }
      ans[i] = ans[i - 1];
    }
    while (!qw[i].empty()) {
      Nig[{i, qw[i].back()}] = (B[qw[i].back()] ? ans[i] : 0) % MD;
      qw[i].pop_back();
    }
  }
  for (int i = 1; i <= q; i++) {
    cout << Nig[{l[i], X[i]}] << '\n';
  }
  return 0;
}
