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 = 1000 * 1000 * 1000 + 7;
const long long MAXN = 1e5 + 7;
const long long MAXV = (1 << 20);
long long a[MAXN];
long long ans[MAXN];
bool dp1[MAXV], dp2[MAXV];
vector<pair<long long, long long> > q[MAXN];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < m; ++i) {
    long long l, x;
    cin >> l >> x;
    q[l - 1].push_back({x, i});
  }
  dp1[0] = 1;
  long long val = 1;
  for (long long i = 0; i < n; ++i) {
    if (dp1[a[i]]) {
      val *= 2;
      val %= MOD;
    } else {
      for (long long j = 0; j < MAXV; ++j) dp2[j] = dp1[j] | dp1[j ^ a[i]];
      for (long long j = 0; j < MAXV; ++j) dp1[j] = dp2[j];
    }
    for (pair<long long, long long> nq : q[i]) {
      ans[nq.second] = val * dp1[nq.first];
    }
  }
  for (long long i = 0; i < m; ++i) {
    cout << ans[i] << '\n';
  }
  return 0;
}
