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;
int pw2[100010];
namespace linear {
class xxj {
 public:
  long long p[65];
  int cnt = 0;
  void clear() { memset(p, 0, sizeof(p)); }
  void insert(long long val) {
    for (int i = 63; i >= 0; i--) {
      if ((val >> i) & 1) {
        if (p[i])
          val ^= p[i];
        else {
          p[i] = val;
          cnt++;
          return;
        }
      }
    }
  }
  long long maxval() {
    long long out = 0;
    for (int i = 63; i >= 0; i--) {
      if ((out ^ p[i]) > out) out ^= p[i];
    }
    return out;
  }
  bool check(int val) {
    for (int i = 63; i >= 0; i--) {
      if ((val >> i) & 1) val ^= p[i];
    }
    return val == 0;
  }
  long long minval() {
    for (int i = 0; i <= 63; i++)
      if (p[i]) return p[i];
    return -1;
  }
};
}  // namespace linear
using namespace linear;
xxj x[100010];
int n, q, l, X;
int main() {
  scanf("%d%d", &n, &q);
  pw2[0] = 1;
  for (int i = 1; i <= n; i++) pw2[i] = (pw2[i - 1] * 2) % 1000000007;
  for (int i = 1, a; i <= n; i++) {
    x[i] = x[i - 1];
    scanf("%d", &a);
    x[i].insert(a);
  }
  while (q--) {
    scanf("%d%d", &l, &X);
    if (!x[l].check(X))
      puts("0");
    else
      printf("%d\n", pw2[l - x[l].cnt]);
  }
  return 0;
}
