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>
long long gcd(long long x, long long y) {
  return y % x == 0 ? x : gcd(y % x, x);
}
template <class T>
T my_abs(T a) {
  if (a < 0) a = -a;
  return a;
}
const int maxn = 1e5 + 7, maxv = (2 << 20) + 1, mod = 1e9 + 7;
int a[maxn], n, q;
std::vector<std::pair<int, int> > v[maxn];
bool have[maxv];
std::vector<int> s;
int res[maxn];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", a + i);
  for (int i = (int)(1); i <= (int)(q); i++) {
    int l, x;
    scanf("%d%d", &l, &x);
    v[l].emplace_back(std::pair<int, int>(x, i));
  }
  s.emplace_back(0);
  have[0] = 1;
  long long ans = 1;
  for (int i = (int)(1); i <= (int)(n); i++) {
    if (have[a[i]])
      ans = ans * 2 % mod;
    else {
      int size = (int)(s).size();
      for (int j = (int)(0); j <= (int)(size - 1); j++) {
        s.emplace_back(s[j] ^ a[i]);
        have[s.back()] = 1;
      }
    }
    for (int j = (int)(0); j <= (int)((int)(v[i]).size() - 1); j++)
      if (have[v[i][j].first]) res[v[i][j].second] = ans;
  }
  for (int i = (int)(1); i <= (int)(q); i++) printf("%d\n", res[i]);
  return 0;
}
