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;
int basis[21];
int sz;
int t[100001];
int mod = 1000000007;
void insert(int msk) {
  int temp = msk;
  for (int i = 0; i < 21; i++) {
    if ((msk & (1 << i)) == 0) continue;
    if (!basis[i]) {
      basis[i] = msk;
      ++sz;
      return;
    }
    msk ^= basis[i];
  }
}
bool query(int num) {
  for (int i = 0; i < 20; i++) {
    if (num & (1 << i)) {
      if (!basis[i]) return 0;
      num ^= basis[i];
    }
  }
  return 1;
}
int main() {
  int n;
  cin >> n;
  int q;
  cin >> q;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<pair<int, int>> queries[n];
  for (int i = 0; i < q; i++) {
    int l, x;
    cin >> l >> x;
    --l;
    queries[l].push_back({x, i});
  }
  t[0] = 1;
  for (int i = 1; i <= n; i++) {
    t[i] = t[i - 1] * 2;
    if (t[i] > mod) t[i] -= mod;
  }
  int ans[q];
  for (int i = 0; i < n; i++) {
    insert(a[i]);
    for (auto &it : queries[i]) {
      if (!query(it.first))
        ans[it.second] = 0;
      else
        ans[it.second] = t[i + 1 - sz];
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << endl;
}
