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> cac;
long long n, q;
vector<pair<long long, long long> > query[100005];
long long a[100005];
long long ans[100005];
queue<long long> temp;
signed main() {
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i <= q; i++) {
    long long l, x;
    cin >> l >> x;
    query[l].push_back({x, i});
  }
  long long get = 1;
  cac.insert(0);
  for (long long i = 1; i <= n; i++) {
    if (!cac.count(a[i])) {
      for (auto j : cac) {
        temp.push(a[i] ^ j);
      }
      while (temp.size()) {
        cac.insert(temp.front());
        temp.pop();
      }
    } else {
      get = get * 2;
      get %= 1000000007;
    }
    for (auto ii : query[i]) {
      if (cac.count(ii.first)) {
        ans[ii.second] = get;
      }
    }
  }
  for (long long i = 1; i <= q; i++) {
    cout << ans[i] << endl;
  }
}
