#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int M = 25;
const int N = 1e5 + 20;
int n, q, tot, cnt;
int a[N], lg[N], ans[N];
struct node {
  int l, x, id;
  bool operator<(const node& a) const {
    if (l != a.l) return l < a.l;
    return x < a.x;
  }
} b[N];
int xxj[M + 5];
void getxxj(int x) {
  for (int i = M; i >= 0; i--) {
    if (x & (1 << i)) {
      if (xxj[i])
        x ^= xxj[i];
      else {
        xxj[i] = x;
        tot++;
        break;
      }
    }
  }
}
bool check(int x) {
  for (int i = M; i >= 0; i--) {
    if (x & (1 << i)) {
      if (xxj[i])
        x ^= xxj[i];
      else
        return 0;
    }
  }
  return 1;
}
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", &b[i].l, &b[i].x);
    b[i].id = i;
  }
  lg[0] = 1;
  for (int i = 1; i <= n; i++) lg[i] = lg[i - 1] * 2 % mod;
  sort(b + 1, b + q + 1);
  for (int i = 1; i <= q; i++) {
    while (cnt < b[i].l) getxxj(a[++cnt]);
    if (check(b[i].x)) ans[b[i].id] = lg[cnt - tot];
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}

### 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;
const int mod = 1e9 + 7;
const int M = 25;
const int N = 1e5 + 20;
int n, q, tot, cnt;
int a[N], lg[N], ans[N];
struct node {
  int l, x, id;
  bool operator<(const node& a) const {
    if (l != a.l) return l < a.l;
    return x < a.x;
  }
} b[N];
int xxj[M + 5];
void getxxj(int x) {
  for (int i = M; i >= 0; i--) {
    if (x & (1 << i)) {
      if (xxj[i])
        x ^= xxj[i];
      else {
        xxj[i] = x;
        tot++;
        break;
      }
    }
  }
}
bool check(int x) {
  for (int i = M; i >= 0; i--) {
    if (x & (1 << i)) {
      if (xxj[i])
        x ^= xxj[i];
      else
        return 0;
    }
  }
  return 1;
}
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", &b[i].l, &b[i].x);
    b[i].id = i;
  }
  lg[0] = 1;
  for (int i = 1; i <= n; i++) lg[i] = lg[i - 1] * 2 % mod;
  sort(b + 1, b + q + 1);
  for (int i = 1; i <= q; i++) {
    while (cnt < b[i].l) getxxj(a[++cnt]);
    if (check(b[i].x)) ans[b[i].id] = lg[cnt - tot];
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}

```