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 int N = 1 << 20;
int n, q, a[N], ans[N];
pair<pair<int, int>, int> Q[N];
long long cnt;
vector<int> xors;
bool have[N];
void add(int i) {
  if (have[a[i]]) {
    cnt = cnt * 2 % 1000000007;
    return;
  }
  int n = ((int)(xors).size());
  for (int j = 0; j <= n - 1; ++j) {
    int x = a[i] ^ xors[j];
    xors.push_back(x);
    have[x] = true;
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &Q[i].first.first, &Q[i].first.second);
    Q[i].second = i;
  }
  xors = {0};
  cnt = have[0] = 1;
  sort(Q + 1, Q + 1 + q);
  int j = 1;
  for (int i = 1; i <= q; ++i) {
    while (j <= Q[i].first.first) add(j++);
    ans[Q[i].second] = have[Q[i].first.second] * cnt;
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
}
