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 MAX_N = 100005;
vector<pair<int, int> > que[MAX_N];
int basis[20];
int cnt;
int t[MAX_N];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  t[0] = 1;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
    t[i + 1] = t[i] * 2 % 1000000007;
  }
  for (int i = 0; i < (int)(20); ++i) {
    basis[i] = -1;
  }
  vector<int> ans(q);
  for (int i = 0; i < (int)(q); ++i) {
    int a, b;
    cin >> a >> b;
    que[a - 1].push_back((pair<int, int>(b, i)));
  }
  for (int i = 0; i < (int)(n); ++i) {
    for (int j = (int)(20) - 1; j >= 0; --j) {
      if ((a[i] >> j) & 1) {
        if (basis[j] < 0) {
          basis[j] = a[i];
          cnt++;
          break;
        } else {
          a[i] ^= basis[j];
        }
      }
    }
    for (auto&(it) : (que[i])) {
      bool ok = true;
      for (int j = (int)(20) - 1; j >= 0; --j) {
        if ((it.first >> j) & 1) {
          if (basis[j] < 0) {
            ok = false;
            ans[it.second] = 0;
            break;
          } else {
            it.first ^= basis[j];
          }
        }
      }
      if (ok) ans[it.second] = t[i + 1 - cnt];
    }
  }
  for (int i = 0; i < (int)(q); ++i) {
    cout << ans[i] << "\n";
  }
  return 0;
}
