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;
struct Query {
  int i, l, x;
  long long ans;
  Query(int i = 0, int l = 0, int x = 0) : i(i), l(l), x(x) { ans = 0; }
};
class Basis {
  int b[20], cnt;

 public:
  Basis() {
    for (int i = 0; i < 20; i++) b[i] = 0;
    cnt = 0;
  }
  void insert(int x) {
    for (int i = 19; i >= 0; i--) {
      if (x & (1 << i)) {
        if (!b[i]) {
          b[i] = x;
          cnt++;
          return;
        } else
          x ^= b[i];
      }
    }
  }
  bool in_span(int x) {
    for (int i = 19; i >= 0; i--) {
      if (x & (1 << i)) {
        if (b[i])
          x ^= b[i];
        else
          return false;
      }
    }
    return true;
  }
  size_t size() { return cnt; }
};
const int MAXN = 1e5 + 20;
const long long P = 1e9 + 7;
int n, qn, a[MAXN];
Query q[MAXN];
long long pow(long long base, long long exp) {
  long long ans = 1, mul = base;
  while (exp) {
    if (exp % 2) ans = ans * mul % P;
    mul = mul * mul % P;
    exp /= 2;
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &qn);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < qn; i++) {
    scanf("%d%d", &q[i].l, &q[i].x);
    q[i].i = i;
  }
  sort(q, q + qn, [](const Query& a, const Query& b) { return a.l < b.l; });
  Basis b;
  int cur = 1;
  for (int i = 0; i < qn; i++) {
    while (cur <= q[i].l) b.insert(a[cur++]);
    if (b.in_span(q[i].x))
      q[i].ans = pow(2, cur - 1 - b.size());
    else
      q[i].ans = 0;
  }
  sort(q, q + qn, [](const Query& a, const Query& b) { return a.i < b.i; });
  for (int i = 0; i < qn; i++) printf("%lld\n", q[i].ans);
}
