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 MOD = 1e9 + 7;
const int MAXN = 100000;
vector<pair<int, int>> qs[MAXN];
int powM(int a, int p) {
  int ans = 1;
  for (int cur = a; p; p >>= 1, cur = 1ll * cur * cur % MOD) {
    if (p & 1) ans = 1ll * ans * cur % MOD;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<int> A(n), ans(q);
  for (int i = 0; i < n; ++i) cin >> A[i];
  for (int i = 0; i < q; ++i) {
    int l, x;
    cin >> l >> x;
    --l;
    qs[l].emplace_back(x, i);
  }
  set<int, greater<int>> S;
  for (int i = 0; i < n; ++i) {
    for (int c : S) {
      if ((c xor A[i]) < A[i]) A[i] = c xor A[i];
    }
    if (A[i]) S.insert(A[i]);
    for (auto [x, q] : qs[i]) {
      for (int c : S) {
        if ((c xor x) < x) x = c xor x;
      }
      if (x == 0)
        ans[q] = powM(2, i + 1 - S.size());
      else
        ans[q] = 0;
    }
  }
  for (int i = 0; i < q; ++i) cout << ans[i] << '\n';
}
