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;
long long int n, q, a[100005];
pair<pair<long long int, long long int>, long long int> qr[100005];
set<long long int> pres;
long long int val;
long long int ans[100005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (long long int i = 1; i <= n; ++i) cin >> a[i];
  for (long long int i = 1; i <= q; ++i) {
    cin >> qr[i].first.first >> qr[i].first.second;
    qr[i].second = i;
  }
  sort(qr + 1, qr + q + 1);
  long long int cur = 0;
  for (long long int i = 1; i <= q; ++i) {
    while (cur <= qr[i].first.first) {
      if (pres.empty()) {
        pres.insert(a[cur]);
        val = 1;
      } else if (pres.find(a[cur]) != pres.end()) {
        val = (val * 2) % 1000000007;
      } else {
        for (long long int v : pres) pres.insert(a[cur] ^ v);
      }
      cur++;
    }
    if (pres.find(qr[i].first.second) != pres.end())
      ans[qr[i].second] = val;
    else
      ans[qr[i].second] = 0;
  }
  for (long long int i = 1; i <= q; ++i) cout << ans[i] << '\n';
}
