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 = 20;
int a[1 << N], ans[1 << N];
vector<pair<int, int>> q[1 << N];
int can(int x, vector<int> &c) {
  for (int i = N - 1; i >= 0; i--)
    if ((x >> i) & 1) {
      if (c[i] < 0) return 0;
      x ^= c[i];
    }
  return !x;
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) {
    int l, x;
    cin >> l >> x;
    q[l].push_back(make_pair(i, x));
  }
  vector<int> c(N, -1);
  int z = 1;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    for (int j = N - 1; j >= 0; j--)
      if ((x >> j) & 1) {
        if (c[j] < 0) {
          c[j] = x;
          break;
        }
        x ^= c[j];
      }
    if (!x) z = (z + z) % 1000000007;
    for (auto [idx, x] : q[i]) {
      ans[idx] = can(x, c) * z;
    }
  }
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
}
