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 = 1e5 + 5;
const int MOD = 1e9 + 7;
inline int msb(int n) {
  int b = 1;
  while (b < 32) {
    n |= n >> b;
    b <<= 1;
  }
  return (n + 1) >> 1;
}
int A[N];
int res[N];
vector<int> bas[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
  }
  vector<int> cur_bas;
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    cur_bas.push_back(A[i]);
    int sz = cur_bas.size();
    for (int i = 0; i < sz - 1; i++) {
      int ms = msb(cur_bas[i]);
      if (cur_bas.back() & ms) cur_bas.back() ^= cur_bas[i];
    }
    if (cur_bas.back() == 0) {
      cur_bas.pop_back();
      ans += ans;
      if (ans >= MOD) ans -= MOD;
    }
    bas[i] = cur_bas;
    res[i] = ans;
  }
  while (q--) {
    int l, x;
    cin >> l >> x;
    cur_bas = bas[l];
    cur_bas.push_back(x);
    int sz = cur_bas.size();
    for (int i = 0; i < sz - 1; i++) {
      int ms = msb(cur_bas[i]);
      if (cur_bas.back() & ms) cur_bas.back() ^= cur_bas[i];
    }
    if (cur_bas.back() == 0) {
      cout << res[l] << "\n";
    } else {
      cout << 0 << "\n";
    }
  }
  return 0;
}
