Problem: We call an array b_1, b_2, …, b_m good, if there exist two indices i < j such that b_i ⋅ b_j is a [perfect square](https://en.wikipedia.org/wiki/Square_number).

Given an array b_1, b_2, …, b_m, in one action you can perform one of the following: 

  * multiply any element b_i by any prime p; 
  * divide any element b_i by prime p, if b_i is divisible by p. 



Let f(b_1, b_2, …, b_m) be the minimum number of actions needed to make the array b good.

You are given an array of n integers a_1, a_2, …, a_n and q queries of the form l_i, r_i. For each query output f(a_{l_i}, a_{l_i + 1}, …, a_{r_i}).

Input

The first line contains two integers n and q (2 ≤ n ≤ 194 598, 1 ≤ q ≤ 1 049 658) — the length of the array and the number of queries.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 5 032 107) — the elements of the array.

Each of the next q lines contains two integers l_i and r_i (1 ≤ l_i < r_i ≤ n) — the parameters of a query.

Output

Output q lines — the answers for each query in the order they are given in the input.

Example

Input

10 10
34 37 28 16 44 36 43 50 22 13
1 3
4 8
6 10
9 10
3 10
8 9
5 6
1 4
1 7
2 6


Output

2
0
1
3
0
1
1
1
0
0

Note

In the first query of the first sample you can multiply second number by 7 to get 259 and multiply the third one by 37 to get 1036. Then a_2 ⋅ a_3 = 268 324 = 518^2.

In the second query subarray is already good because a_4 ⋅ a_6 = 24^2.

In the third query you can divide 50 by 2 to get 25. Then a_6 ⋅ a_8 = 30^2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int MAXN = 200000;
const int MAXQ = 1050000;
const int MAXA = 5033000;
int as[MAXN];
std::vector<std::pair<int, int> > queries[MAXQ];
int ans[MAXQ];
int near[8][MAXA];
int mf[MAXA];
int mfp[MAXA];
int mfr[MAXA];
void precompute_mf() {
  for (int i = 2; i < MAXA; i++) {
    if (!mf[i]) {
      for (int j = i; j < MAXA; j += i) {
        if (!mf[j]) {
          mf[j] = i;
          if (mf[j / i] == i) {
            mfp[j] = mfp[j / i] + 1;
            mfr[j] = mfr[j / i];
          } else {
            mfp[j] = 1;
            mfr[j] = j / i;
          }
        }
      }
    }
  }
}
int factors[1 << 8] = {1};
int pop[1 << 8];
int best[16];
int main() {
  precompute_mf();
  for (int i = 1; i < (1 << 8); i++) {
    pop[i] = pop[i >> 1] + (i & 1);
  }
  int N, Q;
  scanf("%d %d", &N, &Q);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &as[i]);
  }
  for (int i = 0; i < Q; i++) {
    int L, R;
    scanf("%d %d", &L, &R);
    queries[R].emplace_back(L, i);
  }
  for (int i = 1; i <= N; i++) {
    std::vector<int> pfs;
    for (int x = as[i]; x > 1; x = mfr[x]) {
      if (mfp[x] & 1) {
        pfs.push_back(mf[x]);
      }
    }
    for (int k = 0; k < pfs.size(); k++) {
      factors[1 << k] = pfs[k];
    }
    for (int k = 1; k < (1 << pfs.size()); k++) {
      factors[k] = factors[k - (k & -k)] * factors[k & -k];
    }
    for (int k = 0; k < (1 << pfs.size()); k++) {
      for (int j = 0; j < 8; j++) {
        int& x = best[j + pfs.size() - pop[k]];
        x = std::max(x, near[j][factors[k]]);
      }
    }
    for (int k = 0; k < (1 << pfs.size()); k++) {
      int& x = near[pfs.size() - pop[k]][factors[k]];
      x = std::max(x, i);
    }
    for (auto q : queries[i]) {
      int L = q.first;
      for (int j = 0; j < 16; j++) {
        if (L <= best[j]) {
          ans[q.second] = j;
          break;
        }
      }
    }
  }
  for (int i = 0; i < Q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}