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 + 5;
const int MOD = 1e9 + 7;
const int MAXM = 25;
struct Node {
  int id, rr, x;
  Node() = default;
  Node(int a, int b, int c) : id(a), rr(b), x(c) {}
  inline bool operator<(const Node &a) const { return rr < a.rr; }
} q[MAXN];
int n, t, dat[MAXN], a[MAXM], tot, ans[MAXN], expp[MAXN], cur = 1;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
inline void insert(int x) {
  for (int i = 20; i >= 0; i--) {
    if (!(x & (1 << i))) continue;
    if (a[i])
      x ^= a[i];
    else {
      for (int j = 0; j < i; j++)
        if (x & (1 << j)) x ^= a[j];
      for (int j = i + 1; j <= 20; j++)
        if (a[j] & (1 << i)) a[j] ^= x;
      a[i] = x, tot++;
      return;
    }
  }
}
inline bool check(int x) {
  for (int i = 20; i >= 0; i--) {
    if (!(x & (1 << i))) continue;
    if (a[i])
      x ^= a[i];
    else
      return false;
  }
  return true;
}
int main() {
  n = read(), t = read();
  for (int i = 1; i <= n; i++) dat[i] = read();
  for (int i = 1; i <= t; i++) {
    int x = read(), y = read();
    q[i] = Node(i, x, y);
  }
  sort(q + 1, q + t + 1);
  expp[0] = 1;
  for (int i = 1; i <= n; i++) expp[i] = (expp[i - 1] << 1) % MOD;
  for (int i = 1; i <= n; i++) {
    insert(dat[i]);
    while (q[cur].rr == i) {
      if (check(q[cur].x)) ans[q[cur].id] = expp[i - tot];
      cur++;
    }
  }
  for (int i = 1; i <= t; i++) printf("%d\n", ans[i]);
  return 0;
}
