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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
const long long inf64 = 1e18 + 18;
const int inf32 = 15e8;
using namespace std;
const int N = 1 << 21, M = 1e5 + 5;
const long long mod = 1e9 + 7;
bool used[N];
vector<pair<int, int> > quer[N];
vector<int> base;
int ar[M], ans[M];
void solve() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; ++i) scanf("%d", ar + i);
  for (int i = 0, l, x; i < q; ++i) {
    scanf("%d%d", &l, &x);
    quer[--l].push_back({x, i});
  }
  vector<int> base = {0};
  used[0] = true;
  long long cnt = 1;
  for (int i = 0; i < n; ++i) {
    if (!used[ar[i]]) {
      int sz = base.size();
      for (int j = 0; j < sz; ++j) {
        base.push_back(ar[i] ^ base[j]);
        used[base.back()] = true;
      }
    } else
      cnt = cnt * 2 % mod;
    for (auto &x : quer[i]) ans[x.second] = cnt * used[x.first];
  }
  for (int i = 0; i < q; ++i) printf("%d ", ans[i]);
}
int main() {
  solve();
  return 0;
}
