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 N = (1 << 20);
const int mod = 1e9 + 7;
int n, q, a[N];
int ans[N];
bool have[N];
vector<int> v;
vector<pair<pair<int, int>, int> > Q;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= q; ++i) {
    int l, x;
    cin >> l >> x;
    Q.push_back(pair<pair<int, int>, int>(pair<int, int>(l, x), i));
  }
  sort(Q.begin(), Q.end());
  int j = 1;
  v.push_back(0);
  have[0] = 1;
  int res = 1;
  for (int i = 0; i < q; ++i) {
    for (; j <= Q[i].first.first; ++j) {
      if (!have[a[j]]) {
        int sz = v.size();
        for (int k = 0; k < sz; ++k)
          if (!have[v[k] ^ a[j]])
            have[v[k] ^ a[j]] = 1, v.push_back(v[k] ^ a[j]);
      } else
        res = res * 2 % mod;
    }
    ans[Q[i].second] = res * have[Q[i].first.second];
  }
  for (int i = 1; i <= q; ++i) cout << ans[i] << "\n";
}
