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;
template <typename T>
void printvec(vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i] << " ";
  }
  cout << endl;
}
int N = (1 << 20);
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = power(a, b / 2);
  temp = (temp * temp) % (1000 * 1000 * 1000 + 7);
  if (b % 2 == 0)
    return temp;
  else
    return (temp * a) % (1000 * 1000 * 1000 + 7);
}
int main() {
  int ttt = 1;
  while (ttt--) {
    int n, q, i, j, k;
    cin >> n >> q;
    vector<int> inp(n);
    for (i = 0; i < n; i++) {
      cin >> inp[i];
    }
    vector<vector<pair<int, int> > > queryadj(n);
    for (i = 0; i < q; i++) {
      int a, b;
      cin >> a >> b;
      a--;
      queryadj[a].push_back(make_pair(b, i));
    }
    vector<bool> visited(N, false);
    vector<long long> ans(q);
    vector<int> found;
    visited[0] = true;
    found.push_back(0);
    int mul = 0;
    for (i = 0; i < n; i++) {
      if (visited[inp[i]]) {
        mul++;
      } else {
        int sz = found.size();
        for (j = 0; j < sz; j++) {
          int num = found[j] ^ inp[i];
          visited[num] = true;
          found.push_back(num);
        }
      }
      for (j = 0; j < queryadj[i].size(); j++) {
        if (visited[queryadj[i][j].first]) {
          ans[queryadj[i][j].second] = power(2, mul);
        } else {
          ans[queryadj[i][j].second] = 0;
        }
      }
    }
    for (i = 0; i < q; i++) {
      printf("%lld\n", ans[i]);
    }
  }
  return 0;
}
