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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
const int MAXQ = 100000;
const int MOD = 1000000007;
const int MAXBIT = 20;
struct Q {
  int cnt, want, id;
};
bool operator<(const Q &a, const Q &b) { return a.cnt < b.cnt; }
int n, nq;
int a[MAXN];
Q q[MAXQ];
int ans[MAXQ];
int b[MAXN], nb;
int ways;
void init() {
  nb = 0;
  ways = 1;
}
void append(int x) {
  for (int i = (0); i < (nb); ++i)
    if ((x ^ b[i]) < x) x ^= b[i];
  if (x == 0) {
    ways = (long long)ways * 2 % MOD;
    return;
  }
  b[nb++] = x;
}
int solve(int x) {
  for (int i = (0); i < (nb); ++i)
    if ((x ^ b[i]) < x) x ^= b[i];
  return x == 0 ? ways : 0;
}
void run() {
  scanf("%d%d", &n, &nq);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  for (int i = (0); i < (nq); ++i)
    scanf("%d%d", &q[i].cnt, &q[i].want), q[i].id = i;
  sort(q, q + nq);
  int at = 0;
  init();
  for (int i = (0); i < (nq); ++i) {
    while (at < q[i].cnt) append(a[at++]);
    ans[q[i].id] = solve(q[i].want);
  }
  for (int i = (0); i < (nq); ++i) printf("%d\n", ans[i]);
}
int main() {
  run();
  return 0;
}
