#include <bits/stdc++.h>
using namespace std;
int n, m, tot, a[100010], ans[100010];
vector<pair<int, int> > q[100010];
set<int> s;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1, l, x; i <= m; i++) {
    scanf("%d%d", &l, &x);
    q[l].emplace_back(i, x);
  }
  s.insert(0);
  tot = 1;
  for (int i = 1; i <= n; i++) {
    if (s.count(a[i]))
      tot = 1ll * tot * 2 % 1000000007;
    else {
      vector<int> v;
      for (int u : s) v.push_back(u ^ a[i]);
      for (int u : v) s.insert(u);
    }
    for (auto &p : q[i]) ans[p.first] = s.count(p.second) ? tot : 0;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 n, m, tot, a[100010], ans[100010];
vector<pair<int, int> > q[100010];
set<int> s;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1, l, x; i <= m; i++) {
    scanf("%d%d", &l, &x);
    q[l].emplace_back(i, x);
  }
  s.insert(0);
  tot = 1;
  for (int i = 1; i <= n; i++) {
    if (s.count(a[i]))
      tot = 1ll * tot * 2 % 1000000007;
    else {
      vector<int> v;
      for (int u : s) v.push_back(u ^ a[i]);
      for (int u : v) s.insert(u);
    }
    for (auto &p : q[i]) ans[p.first] = s.count(p.second) ? tot : 0;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}

```