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;
#pragma warning(disable : 4996)
const int INF = 0x7fffffff;
const int MAX_N = 1e5 + 5;
int a[MAX_N], q[MAX_N], ask[MAX_N];
const long long mod = 1e9 + 7;
const int K = 20;
long long _2[MAX_N];
struct datastruct {
  int a[K], sz;
  bool insert(int x) {
    for (int i = K - 1; i >= 0; i--)
      if (x >> i & 1) {
        if (!a[i]) {
          a[i] = x;
          return 1;
        }
        x ^= a[i];
      }
    sz++;
    return 0;
  }
  int query_mx() {
    int ans = 0;
    for (int i = K - 1; i >= 0; i--)
      if ((ans ^ a[i]) > ans) ans ^= a[i];
    return ans;
  }
  bool find(int x) {
    for (int i = K - 1; i >= 0; i--)
      if (x >> i & 1) x ^= a[i];
    return x ? 0 : 1;
  }
} T[MAX_N];
int N, Q;
void solve() {
  for (int i = 1; i <= N; i++) {
    T[i] = T[i - 1];
    T[i].insert(a[i]);
  }
  for (int i = 0; i < Q; i++) {
    if (!T[q[i]].find(ask[i]))
      cout << 0 << endl;
    else
      cout << _2[T[q[i]].sz] << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  _2[0] = 1;
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) {
    _2[i] = _2[i - 1] * 2 % mod;
    cin >> a[i];
  }
  for (int i = 0; i < Q; i++) {
    cin >> q[i] >> ask[i];
  }
  solve();
}
