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 mxn = 1e5 + 10;
const int MOD = 1e9 + 7;
int n, q;
long long A[mxn], ans[mxn], two[mxn];
vector<vector<pair<long long, int>>> Q(mxn);
void solve() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  int l, x;
  for (int i = 0; i < q; i++) {
    cin >> l >> x;
    Q[l].push_back({x, i});
  }
  two[0] = (long long)1;
  for (int i = 1; i <= n; i++) {
    two[i] = two[i - 1] << 1;
    if (two[i] > MOD) {
      two[i] %= MOD;
    }
  }
  vector<long long> basis;
  for (int l = 0; l < n; l++) {
    long long a = A[l];
    for (long long b : basis) {
      a = min(a, a ^ b);
    }
    if (a) {
      basis.push_back(a);
    }
    for (auto qry : Q[l + 1]) {
      long long x = qry.first;
      int i = qry.second;
      for (long long b : basis) {
        x = min(x, x ^ b);
      }
      if (x) {
        ans[i] = 0;
      } else {
        ans[i] = two[l + 1 - basis.size()];
      }
    }
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << '\n';
  }
}
int main() { solve(); }
