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 long long N = 200005, INF = 2000000000000000000;
long long power(long long a, long long b, long long p) {
  if (a == 0) return 0;
  long long res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
vector<long long> prime;
bool is_composite[N];
void sieve(long long n) {
  memset(is_composite, false, sizeof(is_composite));
  for (long long i = 2; i < n; ++i) {
    if (!is_composite[i]) prime.push_back(i);
    for (long long j = 0; j < prime.size() && i * prime[j] < n; ++j) {
      is_composite[i * prime[j]] = true;
      if (i % prime[j] == 0) break;
    }
  }
}
void print(bool n) {
  if (n) {
    cout << "YES";
  } else {
    cout << "NO";
  }
}
long long basis[20];
long long upd[N];
long long sz = 0;
long long till = 0;
void insert(long long msk, long long ind) {
  upd[ind] = till;
  for (long long i = 0; i < 20; i++) {
    if (!(msk & 1 << i)) continue;
    if (!basis[i]) {
      till |= (1 << i);
      upd[ind] = till;
      basis[i] = msk;
      return;
    }
    msk ^= basis[i];
  }
}
bool check(long long val, long long ind) {
  for (long long i = 0; i < 20; i++) {
    if (!(val & 1 << i)) continue;
    if ((upd[ind] & 1 << i)) {
      val ^= basis[i];
    } else {
      return false;
    }
  }
  return true;
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, m;
  cin >> n >> m;
  memset(upd, 0, sizeof(upd));
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    insert(x, i);
  }
  while (m--) {
    long long l, x;
    cin >> l >> x;
    if (check(x, l)) {
      long long sz = __builtin_popcountll(upd[l]);
      cout << power(2, l - sz, 1000000007) << "\n";
    } else {
      cout << 0 << "\n";
    }
  }
}
