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 long long N = 1e5 + 5;
const long long MOD = 1e9 + 7;
struct Gauss {
  static const long long bits = 63;
  long long table[bits];
  Gauss() {
    for (long long i = 0; i < bits; i++) table[i] = 0;
  }
  long long size() {
    long long ans = 0;
    for (long long i = 0; i < bits; i++) {
      if (table[i]) ans++;
    }
    return ans;
  }
  bool can(long long x) {
    for (long long i = bits - 1; i >= 0; i--) x = min(x, x ^ table[i]);
    return x == 0;
  }
  void add(long long x) {
    for (long long i = bits - 1; i >= 0 && x; i--) {
      if (table[i] == 0) {
        table[i] = x;
        x = 0;
      } else
        x = min(x, x ^ table[i]);
    }
  }
  long long getBest() {
    long long x = 0;
    for (long long i = bits - 1; i >= 0; i--) x = max(x, x ^ table[i]);
    return x;
  }
};
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
long long n, q;
Gauss store[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    store[i] = store[i - 1];
    store[i].add(x);
  }
  for (long long i = 1; i <= q; i++) {
    long long idx, x;
    cin >> idx >> x;
    if (store[idx].can(x))
      cout << pow(2LL, idx - store[idx].size(), MOD) << "\n";
    else
      cout << "0"
           << "\n";
  }
}
