#include <bits/stdc++.h>
using namespace std;
int has[1200005], cev[1200005 / 10];
int n, q, l, x;
int main() {
  fill(has, has + 1200005, 1005000000);
  scanf("%d %d", &n, &q);
  vector<int> ok;
  has[0] = 0;
  ok.push_back(0);
  int res = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (has[x] == 1005000000) {
      int sz = ((int)ok.size());
      for (int j = 0; j < sz; j++) {
        has[ok[j] ^ x] = i;
        ok.push_back(ok[j] ^ x);
      }
    } else
      res = res * 2 % 1000000007;
    cev[i] = res;
  }
  while (q--) {
    scanf("%d %d", &l, &x);
    printf("%d\n", cev[l] * (has[x] <= l));
  }
}

### Prompt

Please create a solution in Cpp to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int has[1200005], cev[1200005 / 10];
int n, q, l, x;
int main() {
  fill(has, has + 1200005, 1005000000);
  scanf("%d %d", &n, &q);
  vector<int> ok;
  has[0] = 0;
  ok.push_back(0);
  int res = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (has[x] == 1005000000) {
      int sz = ((int)ok.size());
      for (int j = 0; j < sz; j++) {
        has[ok[j] ^ x] = i;
        ok.push_back(ok[j] ^ x);
      }
    } else
      res = res * 2 % 1000000007;
    cev[i] = res;
  }
  while (q--) {
    scanf("%d %d", &l, &x);
    printf("%d\n", cev[l] * (has[x] <= l));
  }
}

```