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 mxN = 1e5 + 10, N = 20, mod = 1e9 + 7;
vector<int> adj[mxN];
pair<int, int> l[mxN];
int power[mxN], basis[N + 1], sz = 0, ar[mxN], ans[mxN];
void insertVector(int mask) {
  for (int i = 0; i < N; i++) {
    if ((mask & 1 << i) == 0) continue;
    if (!basis[i]) {
      basis[i] = mask;
      sz++;
      return;
    }
    mask ^= basis[i];
  }
}
bool checkXor(int mask) {
  for (int i = 0; i <= N; i++) {
    if ((mask & 1 << i) == 0) continue;
    if (!basis[i]) return false;
    mask ^= basis[i];
  }
  return true;
}
int main() {
  int n, m, i, j, a, b;
  cin >> n >> m;
  for (i = 1; i <= n; i++) cin >> ar[i];
  for (i = 1; i <= m; i++)
    cin >> l[i].first >> l[i].second, adj[l[i].first].push_back(i);
  power[0] = 1;
  for (i = 1; i < mxN; i++) power[i] = (power[i - 1] * 2LL) % mod;
  for (i = 1; i <= n; i++) {
    insertVector(ar[i]);
    for (auto cur : adj[i])
      if (checkXor(l[cur].second)) ans[cur] = power[i - sz];
  }
  for (i = 1; i <= m; i++) cout << ans[i] << endl;
  return 0;
}
