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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && (c != '-')) c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const int N = 2e5 + 10, M = 1050010;
int n, q, a[N], ans[M];
vector<pair<int, int> > Q[N];
vector<int> b[N];
const int maxn = 5050000 + 10;
int mn[maxn], cnt, p[maxn], l[maxn][8];
inline void init(int n) {
  for (register int i = (2); i <= (n); i++) {
    if (!mn[i]) p[++cnt] = i, mn[i] = i;
    for (int j = 1; j <= cnt && i * p[j] <= n; ++j) {
      mn[i * p[j]] = mn[i];
      if (i % p[j] == 0) break;
    }
  }
}
inline vector<int> get(int x) {
  vector<int> ret;
  while (x > 1) {
    int cnt = 0, t = mn[x];
    while (mn[x] == t) x /= mn[x], cnt++;
    if (cnt & 1) ret.push_back(t);
  }
  return ret;
}
const int inf = 1 << 30;
int c[N];
inline void Clear() {
  for (register int i = (1); i <= (n); i++) c[i] = inf;
}
inline void Add(int x, int y) {
  for (; x; x -= x & -x) c[x] = min(c[x], y);
}
inline int Query(int x) {
  int ret = inf;
  for (; x <= n; x += x & -x) ret = min(ret, c[x]);
  return ret;
}
int main() {
  n = read(), q = read(), init(maxn - 10);
  for (register int i = (1); i <= (n); i++) {
    a[i] = read();
    b[i] = get(a[i]);
  }
  for (register int i = (1); i <= (q); i++) {
    int l = read(), r = read();
    Q[r].emplace_back(l, i);
  }
  Clear();
  for (register int i = (1); i <= (n); i++) {
    for (register int j = (0); j < (1 << ((int)(b[i]).size())); j++) {
      int x = 1, tot = 0;
      for (register int k = (0); k < (((int)(b[i]).size())); k++)
        if (j & (1 << k)) x *= b[i][k], ++tot;
      for (register int k = (0); k <= (7); k++)
        if (l[x][k]) {
          int y = k + ((int)(b[i]).size()) - tot;
          Add(l[x][k], y);
        }
      l[x][((int)(b[i]).size()) - tot] = i;
    }
    for (auto j : Q[i]) ans[j.second] = Query(j.first);
  }
  for (register int i = (1); i <= (q); i++) printf("%d\n", ans[i]);
}