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 = 100005;
const int mod = 1e9 + 7;
int n, q, a[N];
vector<pair<int, int> > queries[N];
bool have[1 << 20];
int res[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 0; i < q; ++i) {
    int l, x;
    cin >> l >> x;
    queries[l].push_back(make_pair(i, x));
  }
  have[0] = 1;
  vector<int> vec(1, 0);
  int cur = 1;
  for (int i = 1; i <= n; ++i) {
    if (have[a[i]]) {
      cur = 2LL * cur % mod;
    } else {
      int sz = vec.size();
      for (int j = 0; j < sz; ++j) {
        vec.push_back(vec[j] ^ a[i]);
        have[vec[j] ^ a[i]] = true;
      }
    }
    for (auto &it : queries[i]) {
      int id = it.first;
      int x = it.second;
      if (have[x]) {
        res[id] = cur;
      } else {
        res[id] = 0;
      }
    }
  }
  for (int i = 0; i < q; ++i) {
    printf("%d\n", res[i]);
  }
}
