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 N = 200031;
vector<int> entries[N];
int l[N], x[N];
int n, tests;
int ar[N];
vector<int> basis, temp_basis;
int ans[N];
long long pw(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2) return a * pw(a, b - 1) % 1000000007;
  return pw(a * a % 1000000007, b / 2);
}
int get_high(int val) {
  for (int i = 20; i >= 0; --i) {
    if (val & (1 << i)) return i;
  }
  return -1;
}
vector<int> add_basis(vector<int> basis, int val) {
  for (int i = 0; i < basis.size(); i++) {
    int Q = get_high(basis[i]);
    if (val & (1 << Q)) {
      val ^= basis[i];
    }
  }
  if (val != 0) {
    basis.push_back(val);
    sort(basis.begin(), basis.end());
    reverse(basis.begin(), basis.end());
  }
  return basis;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> tests;
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
  }
  for (int test = 1; test <= tests; test++) {
    cin >> l[test] >> x[test];
    entries[l[test]].push_back(test);
  }
  for (int i = 1; i <= n; i++) {
    basis = add_basis(basis, ar[i]);
    for (int j = 0; j < entries[i].size(); j++) {
      int test_id = entries[i][j];
      temp_basis = add_basis(basis, x[test_id]);
      if (temp_basis.size() != basis.size()) {
        ans[test_id] = 0;
      } else {
        ans[test_id] = pw(2, i - basis.size());
      }
    }
  }
  for (int i = 1; i <= tests; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}
