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 inf = 1e18;
const int M = 1e9 + 7;
string to_string(int x) {
  stringstream second;
  second << x;
  return second.str();
}
int to_integer(string s) {
  stringstream second(s);
  int x;
  second >> x;
  return x;
}
vector<pair<int, int> > p[100005];
vector<int> v;
int dp[100005], sel[2000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, q;
  cin >> n >> q;
  int a[n + 1];
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= q; i++) {
    int l, x;
    cin >> l >> x;
    p[l].push_back({x, i});
  }
  v.push_back(0);
  sel[0] = 1;
  int val = 1;
  for (int i = 1; i <= n; i++) {
    if (sel[a[i]])
      val = (val * 2) % M;
    else {
      int tm = v.size();
      for (int j = 0; j < tm; j++) {
        int x = v[j] ^ a[i];
        v.push_back(x);
        sel[x] = 1;
      }
    }
    for (int j = 0; j < p[i].size(); j++) {
      dp[p[i][j].second] = val * (sel[p[i][j].first]);
    }
  }
  for (int i = 1; i <= q; i++) cout << dp[i] << "\n";
  return 0;
}
