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>
const int mod = 1000000007;
const int inf = 1000000009;
const long long INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.0000000001;
using namespace std;
int B[100005][21];
int S[100005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n, q;
  cin >> n >> q;
  S[0] = 1;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    for (int j = 19; j >= 0; j--) B[i][j] = B[i - 1][j];
    bool czy = false;
    for (int j = 19; j >= 0; j--)
      if (a & (1 << j)) {
        if (!B[i][j]) {
          czy = true;
          B[i][j] = a;
          break;
        }
        a ^= B[i][j];
      }
    S[i] = S[i - 1];
    if (!czy) {
      S[i] *= 2;
      S[i] %= mod;
    }
  }
  for (int i = 1; i <= q; i++) {
    int l, a;
    cin >> l >> a;
    bool czy = false;
    for (int j = 19; j >= 0; j--) {
      if (a & (1 << j)) {
        if (!B[l][j]) {
          czy = true;
          break;
        }
        a ^= B[l][j];
      }
    }
    if (!czy)
      cout << S[l] << "\n";
    else
      cout << "0\n";
  }
  return 0;
}
