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 = 100010;
int b[N], res[N];
vector<pair<int, int> > a[N];
bool vis[(1 << 20) + 10];
vector<int> s;
int main() {
  int n, q, x, y, z;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
  }
  for (int i = 1; i <= q; i++) {
    cin >> x >> y;
    a[x].push_back({i, y});
  }
  long long r = 1;
  s.push_back(0);
  vis[0] = 1;
  for (int i = 1; i <= n; i++) {
    if (vis[b[i]]) {
      r = (2 * r) % 1000000007;
    } else {
      int y = s.size();
      for (int x = 0; x < y; x++)
        s.push_back(s[x] ^ b[i]), vis[s[x] ^ b[i]] = 1;
      s.push_back(b[i]);
      vis[b[i]] = 1;
    }
    for (auto x : a[i])
      if (vis[x.second]) res[x.first] = r;
  }
  for (int i = 1; i <= q; i++) cout << res[i] << " ";
  return 0;
}
