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 + 10, maxq = 1.1e6, maxa = 5.1e6;
inline char gc() {
  return getchar();
  static const int maxs = 1 << 16;
  static char buf[maxs], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, maxs, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long rd() {
  long long x = 0;
  char c = gc();
  bool neg = 0;
  while (c < '0' || c > '9') {
    if (c == '-') neg = 1;
    c = gc();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = gc();
  return neg ? (~x + 1) : x;
}
int a[maxn], N, Q;
struct Node {
  int l, i;
};
vector<Node> que[maxn];
int ans[maxq], f[maxa][10], rp[15];
bool notp[maxa];
int pri[2333333], cnt;
int main() {
  int i, j, k;
  N = rd(), Q = rd();
  for (i = 1; i <= N; i++) a[i] = rd();
  for (i = 1; i <= Q; i++) {
    Node x = (Node){(int)rd(), i};
    que[rd()].push_back(x);
  }
  notp[0] = notp[1] = 1;
  for (i = 2; i <= 5032107; i++) {
    if (!notp[i]) pri[++cnt] = i;
    for (j = 1; j <= cnt && 1ll * pri[j] * i <= 5032107; j++) {
      notp[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
    }
  }
  for (i = 1; i <= N; i++) {
    int p[10], n = 0;
    for (j = 1; pri[j] * pri[j] <= a[i] && j <= cnt; j++) {
      int cc = 0;
      while (a[i] % pri[j] == 0) a[i] /= pri[j], cc++;
      if (cc & 1) p[++n] = pri[j];
    }
    if (a[i] != 1) p[++n] = a[i];
    for (j = 0; j < (1 << n); j++) {
      int g = 1, m = n;
      for (k = 1; k <= n; k++) {
        if ((j >> (k - 1)) & 1) g *= p[k], m--;
      }
      for (k = 0; k <= 7; k++) {
        rp[k + m] = max(rp[k + m], f[g][k]);
      }
      f[g][m] = i;
    }
    for (j = 0; j < que[i].size(); j++) {
      for (k = 0; k <= 14; k++) {
        if (rp[k] >= que[i][j].l) {
          ans[que[i][j].i] = k;
          break;
        }
      }
    }
  }
  for (i = 1; i <= Q; i++) printf("%d\n", ans[i]);
  return 0;
}