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 = 1 << 20, MOD = 1000 * 1000 * 1000 + 7;
int n, q, dex, x, now, t[N], cnt[N], power[N];
bool have[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  power[0] = 1;
  for (int i = 1; i < N; i++) {
    power[i] = power[i - 1] << 1;
    if (power[i] >= MOD) power[i] -= MOD;
  }
  cin >> n >> q;
  t[0] = 1;
  have[0] = true;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (have[x]) {
      cnt[i] = now;
      continue;
    }
    cnt[i] = ++now;
    for (int j = 0; j < N; j++)
      if (have[j] == false && have[j ^ x]) {
        have[j] = true;
        t[j] = i;
      }
  }
  while (q--) {
    cin >> dex >> x;
    cout << (t[x] && t[x] <= dex ? power[dex - cnt[dex]] : 0) << '\n';
  }
  return 0;
}
