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 MN = 200005, MX = 5100005, MQ = 1100005, MD = 8, inf = 1000000005,
          MOD = 1000000007, P2 = (1 << 18), MTS = 2 * P2 + 5;
const long long INF = 1000000000000000005LL;
int dam[MX][MD];
int naj[MX], ans[MQ];
vector<pair<int, int> > Q[MN];
int co[MN];
int T[MTS];
void prep() {
  for (int i = 2; i < MX; ++i)
    for (int j = i; j <= MX; j += i)
      if (!naj[j]) naj[j] = i;
  for (int i = 0; i < MTS; ++i) T[i] = inf;
}
inline void add(int x, int p) {
  p += P2;
  while (p) {
    T[p] = min(T[p], x);
    p >>= 1;
  }
}
inline int ask(int a, int b) {
  a += P2, b += P2;
  int ret = inf;
  while (a < b) {
    if (a & 1) ret = min(ret, T[a++]);
    if (!(b & 1)) ret = min(ret, T[b--]);
    a >>= 1, b >>= 1;
  }
  if (a == b) ret = min(ret, T[a]);
  return ret;
}
int main() {
  prep();
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &co[i]);
  for (int i = 1; i <= q; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    Q[a].push_back({b, i});
  }
  vector<int> pri;
  for (int i = n; i > 0; --i) {
    pri.clear();
    int v = co[i], cp, pa;
    while (v > 1) {
      cp = naj[v], pa = 0;
      while (v % cp == 0) {
        v /= cp;
        pa ^= 1;
      }
      if (pa == 1) pri.push_back(cp);
    }
    int x = pri.size();
    int p2 = (1 << x);
    for (int j = 0; j < p2; ++j) {
      int cur = 1, moje = x - __builtin_popcount(j);
      for (int k = 0; k < x; ++k)
        if ((j >> k) & 1) cur *= pri[k];
      for (int k = 0; k <= 7; ++k)
        if (dam[cur][k] != 0) add(k + moje, dam[cur][k]);
      dam[cur][moje] = i;
    }
    for (auto x : Q[i]) ans[x.second] = ask(i, x.first);
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", ans[i]);
}