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;
set<long long int> s;
long long int a[100005];
long long int v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, q;
  cin >> n >> q;
  v = 1;
  s.insert(0);
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  pair<long long int, long long int> p[q + 1];
  long long int ans[q + 1];
  p[q].first = -1;
  for (long long int i = 0; i < q; i++) {
    cin >> p[i].first >> ans[i];
    p[i].second = i;
  }
  sort(p, p + q);
  s.clear();
  s.insert(0);
  long long int y = 0;
  for (long long int i = 1; i <= n; i++) {
    auto x = s.find(a[i]);
    if (x == s.end()) {
      for (auto j : s) {
        s.insert((a[i] ^ j));
      }
    } else {
      v *= 2;
      v %= 1000000007;
    }
    while (p[y].first == i) {
      x = s.find(ans[p[y].second]);
      if (x != s.end())
        ans[p[y].second] = v;
      else
        ans[p[y].second] = 0;
      y++;
    }
  }
  for (long long int i = 0; i < q; i++) {
    cout << ans[i] << '\n';
  }
}
