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 = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int N = 1e5 + 5;
const long long mod = 1e9 + 7;
int n, q, a[N];
unordered_set<int> s;
long long dp[N], res[N];
int main(void) {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<tuple<int, int, int> > vq;
  for (int i = 0; i < q; i++) {
    int l, x;
    cin >> l >> x;
    vq.emplace_back(l, x, i);
  }
  sort(vq.begin(), vq.end());
  int i = 0;
  long long val = 1;
  s.insert(0);
  for (auto it : vq) {
    int l, x, id;
    tie(l, x, id) = it;
    while (i < l) {
      if (s.count(a[i]))
        val = (val << 1) % mod;
      else {
        vector<int> add;
        for (auto it : s) add.push_back(it ^ a[i]);
        for (auto it : add) s.insert(it);
      }
      i++;
    }
    if (s.count(x)) res[id] = val;
  }
  for (int i = 0; i < q; i++) cout << res[i] << endl;
  return 0;
}
