#include <bits/stdc++.h>
using namespace std;
struct Node {
  int l, x, id;
};
const int N = 1e5 + 10;
const long long mod = 1e9 + 7;
int d[21], a[N];
long long pow2[N], ans[N];
Node b[N];
int tot;
void insert(int x) {
  for (int i = 20; ~i; i--) {
    if (x & (1 << i)) {
      if (d[i])
        x ^= d[i];
      else {
        d[i] = x;
        tot++;
        return;
      }
    }
  }
}
int query(int x = 0) {
  int ans = x;
  for (int i = 20; ~i; i--) {
    if ((ans ^ d[i]) < ans) ans ^= d[i];
  }
  return ans;
}
bool cmp(const Node& x, const Node& y) { return x.l < y.l; }
int main() {
  int n, q, p;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  pow2[0] = 1;
  for (int i = 1; i <= n; i++) pow2[i] = pow2[i - 1] * 2 % mod;
  for (int i = 1; i <= q; i++) {
    scanf("%d %d", &b[i].l, &b[i].x);
    b[i].id = i;
  }
  sort(b + 1, b + q + 1, cmp);
  p = 0;
  tot = 0;
  for (int i = 1; i <= q; i++) {
    while (p < b[i].l) insert(a[++p]);
    if (query(b[i].x))
      ans[b[i].id] = 0;
    else
      ans[b[i].id] = pow2[b[i].l - tot];
  }
  for (int i = 1; i <= q; i++) printf("%I64d\n", ans[i]);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
struct Node {
  int l, x, id;
};
const int N = 1e5 + 10;
const long long mod = 1e9 + 7;
int d[21], a[N];
long long pow2[N], ans[N];
Node b[N];
int tot;
void insert(int x) {
  for (int i = 20; ~i; i--) {
    if (x & (1 << i)) {
      if (d[i])
        x ^= d[i];
      else {
        d[i] = x;
        tot++;
        return;
      }
    }
  }
}
int query(int x = 0) {
  int ans = x;
  for (int i = 20; ~i; i--) {
    if ((ans ^ d[i]) < ans) ans ^= d[i];
  }
  return ans;
}
bool cmp(const Node& x, const Node& y) { return x.l < y.l; }
int main() {
  int n, q, p;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  pow2[0] = 1;
  for (int i = 1; i <= n; i++) pow2[i] = pow2[i - 1] * 2 % mod;
  for (int i = 1; i <= q; i++) {
    scanf("%d %d", &b[i].l, &b[i].x);
    b[i].id = i;
  }
  sort(b + 1, b + q + 1, cmp);
  p = 0;
  tot = 0;
  for (int i = 1; i <= q; i++) {
    while (p < b[i].l) insert(a[++p]);
    if (query(b[i].x))
      ans[b[i].id] = 0;
    else
      ans[b[i].id] = pow2[b[i].l - tot];
  }
  for (int i = 1; i <= q; i++) printf("%I64d\n", ans[i]);
  return 0;
}

```