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 size_t MAX = 1e5, MAX_A = 1 << 20, BITS = 20;
const uint32_t MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  uint32_t n, q;
  cin >> n >> q;
  static uint32_t A[MAX], R[MAX];
  for (uint32_t i = 0; i < n; i++) cin >> A[i];
  static tuple<uint32_t, uint32_t, uint32_t> D[MAX];
  for (uint32_t i = 0; i < q; i++)
    cin >> get<0>(D[i]) >> get<1>(D[i]), get<2>(D[i]) = i;
  static bitset<MAX_A> Q;
  vector<uint32_t> S = {0};
  Q[0] = true;
  sort(D, D + q);
  uint32_t iq = 0;
  uint32_t m = 1;
  for (uint32_t i = 0; i < n; i++) {
    if (not Q[A[i]]) {
      if (i == 0 or not Q[A[i] ^ A[i - 1]])
        for (uint32_t j = S.size(); j-- > 0;) {
          S.push_back(S[j] ^ A[i]);
          Q[S[j] ^ A[i]] = true;
        }
    } else
      m *= 2;
    m %= MOD;
    while (iq < q and get<0>(D[iq]) <= i + 1) {
      if (Q[get<1>(D[iq])]) R[get<2>(D[iq])] = m;
      iq++;
    }
  }
  for (uint32_t i = 0; i < q; i++) cout << R[i] << "\n";
}
