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 MOD = 1e9 + 7;
const long long INF = 1e18;
const long double E = 1e-8;
mt19937_64 rnd(time(0));
long long p1 = 31, p2 = 37, p3 = 7, p4 = rnd() % 100 + 1, md1 = MOD,
          md2 = 998244353, md3 = 998244357, md4 = rnd() % 1000000000;
long long n, m, k, h, T, kol = 0, sz = 0, ans[1000001], a[1000001];
long long binpow(long long h, long long r, long long md = MOD) {
  long long l = 1;
  while (r) {
    if (r & 1) l *= h, l %= md;
    h *= h;
    h %= md;
    r /= 2;
  }
  return l;
}
long long gcd(long long x, long long y) {
  if (x < y) swap(x, y);
  while (x && y) {
    x %= y;
    swap(x, y);
  }
  return x + y;
}
long long basis[21];
void insert_into_basis(int h) {
  for (int i = 0; i < 20; i++)
    if (h & (1 << i)) {
      if (basis[i] == 0) {
        basis[i] = h;
        sz++;
        return;
      }
      h ^= basis[i];
    }
}
bool check(int h) {
  for (int i = 0; i < 20; i++)
    if (h & (1 << i)) {
      if (basis[i] == 0) {
        return 0;
      }
      h ^= basis[i];
    }
  return 1;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<pair<long long, pair<long long, long long> > > v;
  for (int i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    v.push_back(make_pair(x, make_pair(y, i)));
  }
  sort(v.begin(), v.end());
  long long l = 0;
  for (int i = 1; i <= n; i++) {
    insert_into_basis(a[i]);
    while (l < v.size() && v[l].first <= i) {
      if (check(v[l].second.first)) {
        ans[v[l].second.second] = binpow(2, i - sz);
      }
      l++;
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << "\n";
  return 0;
}
