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 maxn = 2e5 + 5;
const int maxm = 5032108;
template <class T>
inline bool In(T &ret) {
  char c = getchar();
  ret = 0;
  if (c == EOF) return false;
  T flag = 1;
  while (c != '-' && !isdigit(c)) c = getchar();
  if (c == '-') flag = -1, c = getchar();
  while (isdigit(c)) ret = ret * 10 + (c - '0'), c = getchar();
  ret *= flag;
  return true;
}
void fmax(int &x, int y) {
  if (x < y) x = y;
}
int pri[maxn], cnt = 0;
void GetPrime() {
  for (int i = 2; i < maxn; ++i) {
    if (!pri[i]) pri[cnt++] = i;
    for (int j = 0; j < cnt && pri[j] * i < maxn; ++j) {
      pri[pri[j] * i] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
vector<int> fac[maxn];
vector<pair<int, int> > qr[maxn];
int f[20], d[maxm][8], ans[1049659];
int bit[256], pre[256], w[256];
int main() {
  int n, m, q, i, j, k, x, t;
  GetPrime();
  In(n), In(q);
  for (i = 1; i <= n; i++) {
    In(x);
    for (j = 0; pri[j] * pri[j] <= x; j++) {
      for (t = 0; x % pri[j] == 0; t ^= 1) x /= pri[j];
      if (t) fac[i].push_back(pri[j]);
    }
    if (x > 1) fac[i].push_back(x);
  }
  for (i = 0; i < q; i++) {
    In(x), In(t);
    qr[t].push_back(pair<int, int>(x, i));
  }
  for (i = 1; i < 256; i++) {
    pre[i] = __builtin_ctz(i);
    bit[i] = bit[i ^ (1 << pre[i])] + 1;
  }
  w[0] = 1;
  for (i = 1; i <= n; i++) {
    m = fac[i].size();
    for (j = 0; j < (1 << m); j++) {
      if (j) w[j] = w[j ^ (1 << pre[j])] * fac[i][pre[j]];
      x = m - bit[j];
      for (k = 0; k < 8; k++) fmax(f[k + x], d[w[j]][k]);
    }
    for (auto e : qr[i]) {
      for (k = 0; f[k] < e.first; k++)
        ;
      ans[e.second] = k;
    }
    for (j = 0; j < (1 << m); j++) fmax(d[w[j]][m - bit[j]], i);
  }
  for (i = 0; i < q; i++) printf("%d\n", ans[i]);
  ;
  return 0;
}