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 long long MOD = 1e9 + 7;
long long mul(long long a, long long b) { return (a * b) % MOD; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, q;
  cin >> n >> q;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<pair<long long, long long> > > queries(n);
  for (long long i = 0; i < q; i++) {
    long long l, x;
    cin >> l >> x;
    l--;
    queries[l].push_back({x, i});
  }
  vector<long long> ans(q, 0);
  set<long long> s;
  s.insert(0);
  long long cnt = 1;
  for (long long i = 0; i < n; i++) {
    if (s.find(a[i]) != s.end())
      cnt = mul(cnt, 2);
    else {
      vector<long long> v;
      for (long long x : s) v.push_back(x ^ a[i]);
      for (long long x : v) s.insert(x);
      s.insert(a[i]);
      if (a[i] == 0) cnt = mul(cnt, 2);
    }
    for (pair<long long, long long> query : queries[i]) {
      if (s.find(query.first) != s.end()) ans[query.second] = cnt;
    }
  }
  for (long long x : ans) cout << x << '\n';
  return 0;
}
