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;
vector<int> basis;
long long mod = 1000000007;
long long powmod(int a, int b, long long mod) {
  if (b == 0) return 1;
  long long x = powmod(a, b / 2, mod);
  x = (x * x) % mod;
  if (b % 2) x = (x * a) % mod;
  return x;
}
bool willAdd(int x) {
  for (int t : basis) {
    if ((x ^ t) < x) x ^= t;
  }
  return x != 0;
}
void add(int x) {
  for (int t : basis) {
    if ((x ^ t) < x) x ^= t;
  }
  if (x == 0) return;
  for (int& t : basis) {
    if ((x ^ t) < t) t ^= x;
  }
  basis.push_back(x);
  sort(basis.begin(), basis.end(), greater<int>());
}
long long ans[101010];
vector<pair<int, int> > v[101010];
int a[101010];
int main() {
  int n, q, i;
  cin >> n >> q;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    v[x - 1].push_back({y, i});
  }
  for (i = 0; i < n; i++) {
    add(a[i]);
    for (auto q : v[i]) {
      int id = q.second;
      int x = q.first;
      if (!willAdd(x)) {
        ans[id] = powmod(2, i + 1 - basis.size(), mod);
      }
    }
  }
  for (i = 0; i < q; i++) {
    cout << ans[i] << '\n';
  }
  return 0;
}
