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 maxn = 1e5, mod = 1e9 + 7;
int a[maxn + 5];
int f[maxn + 5];
int base[20];
int ans[maxn + 5];
int n, q;
struct wjmzbmr {
  int l, x, id;
  void input() { scanf("%d%d", &l, &x); }
  bool operator<(const wjmzbmr &a) const { return l < a.l; }
} b[maxn + 5];
void ins(int x) {
  for (int i = 19; i >= 0; --i)
    if (x & (1 << i))
      if (!base[i]) {
        base[i] = x;
        for (int j = 19; j > i; --j)
          if (base[j] & (1 << i)) base[j] ^= base[i];
        break;
      } else
        x ^= base[i];
}
int query(int len, int x) {
  int now = 0;
  for (int i = 19; i >= 0; --i) {
    int p = (x & (1 << i)) > 0;
    int q = (now & (1 << i)) > 0;
    if (p != q)
      if (!base[i])
        return 0;
      else
        now ^= base[i];
  }
  int cnt = 0;
  for (int i = 0; i < 20; ++i)
    if (base[i]) ++cnt;
  return f[len - cnt];
}
int main() {
  f[0] = 1;
  for (int i = 1; i <= maxn; ++i) f[i] = f[i - 1] * 2 % mod;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= q; ++i) b[i].input(), b[i].id = i;
  sort(b + 1, b + q + 1);
  int now = 1;
  for (int i = 1; i <= n; ++i) {
    ins(a[i]);
    for (; now <= q && b[now].l == i; ++now)
      ans[b[now].id] = query(i, b[now].x);
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
  return 0;
}
