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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
using namespace std;
const long long mod = 1e9 + 7;
const long long nmax = 100005;
long long n, q, pute[nmax], a[nmax], curr[21][nmax];
vector<bitset<20>> matrix(nmax);
vector<long long> where(21, -1), update;
void Algo() {
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= 20; j++) {
      if (where[j] != -1 && matrix[i][j] == 1) {
        matrix[i] ^= matrix[where[j]];
      }
    }
    for (long long j = 20; j >= 0; j--) {
      if (matrix[i][j] == 1) {
        where[j] = i;
        update.push_back(i - 1);
        for (long long t = 1; t <= n; t++) {
          for (long long p = 0; p <= 20; p++) {
            curr[(long long)update.size() - 1][t] += (1LL << p) * matrix[t][p];
          }
        }
        for (long long t = 1; t < i; t++) {
          if (matrix[t][j] == 1) {
            matrix[t] ^= matrix[i];
          }
        }
        break;
      }
    }
  }
  update.push_back(n);
  for (long long j = 0; j <= 20; j++) {
    for (long long i = 1; i <= n; i++) {
      curr[(long long)update.size() - 1][i] += (1LL << j) * matrix[i][j];
    }
  }
}
void precalc() {
  pute[0] = 1;
  for (long long i = 1; i < nmax; i++) {
    pute[i] = 1LL * 2 * pute[i - 1] % mod;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j <= 20; j++) {
      if (a[i] & (1LL << j)) matrix[i][j] = 1;
    }
  }
}
long long query(long long l, long long x) {
  long long current = 0, indx = 0;
  for (long long i = 0; i < (long long)update.size(); i++) {
    current = update[i];
    if (current >= l) {
      indx = i;
      break;
    }
  }
  long long ans = 0;
  for (long long j = 0; j <= 20; j++) {
    if (x & (1LL << j)) {
      if (where[j] != -1 && where[j] <= l) {
        ans ^= curr[indx][where[j]];
      }
    }
  }
  if (ans != x)
    return 0;
  else {
    long long cnt = 0;
    for (long long j = 0; j <= 20; j++) {
      if (where[j] != -1 && where[j] <= l) cnt++;
    }
    return pute[l - cnt];
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  srand(chrono::steady_clock::now().time_since_epoch().count());
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  precalc();
  Algo();
  for (long long i = 1; i <= q; i++) {
    long long l, x;
    cin >> l >> x;
    cout << query(l, x) << '\n';
  }
}
