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>
using namespace std;
const int mod = 1e9 + 7;
long long qpow(long long x, long long k) {
  return k == 0 ? 1
                : 1ll * qpow(1ll * x * x % mod, k >> 1) * (k & 1 ? x : 1) % mod;
}
const int Maxn = 2e5 + 5;
const int Maxq = 11e5 + 5;
const int Maxa = 5032107 + 5;
vector<int> prime;
int lp[Maxa];
int simp[Maxa];
int num[Maxa];
int n, q;
int a[Maxn];
int ans[Maxq];
vector<pair<int, int> > Q[Maxn];
int dp[Maxa][8];
int best[15];
void pre() {
  for (int i = 2; i < Maxa; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      prime.push_back(i);
    }
    for (int j = 0; j < prime.size(); j++) {
      if (prime[j] * i >= Maxa) break;
      if (prime[j] > lp[i]) break;
      lp[prime[j] * i] = prime[j];
    }
  }
  simp[1] = 1;
  for (int i = 2; i < Maxa; i++) {
    int cur = i / lp[i];
    if (cur % lp[i] == 0) {
      simp[i] = simp[cur / lp[i]];
      num[i] = num[cur / lp[i]];
    } else {
      simp[i] = simp[cur] * lp[i];
      num[i] = num[cur] + 1;
    }
  }
}
vector<int> divs;
void get_divisor(int x) {
  divs.clear();
  divs.push_back(1);
  while (x > 1) {
    int sz = divs.size();
    for (int i = 0; i < sz; i++) {
      divs.push_back(divs[i] * lp[x]);
    }
    x /= lp[x];
  }
}
int main() {
  pre();
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    a[i] = simp[a[i]];
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    Q[r].push_back(pair<int, int>(l, i));
  }
  for (int i = 1; i <= n; i++) {
    get_divisor(a[i]);
    for (auto d : divs) {
      int add1 = num[a[i]] - num[d];
      for (int add2 = 0; add2 <= 7; add2++) {
        int l = dp[d][add2];
        if (l > best[add1 + add2]) best[add1 + add2] = l;
      }
    }
    for (auto qr : Q[i]) {
      for (int i = 0; i <= 14; i++) {
        if (best[i] >= qr.first) {
          ans[qr.second] = i;
          break;
        }
      }
    }
    for (auto d : divs) {
      int add = num[a[i]] - num[d];
      if (i > dp[d][add]) dp[d][add] = i;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}