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 MAXAI = 5032107, MAXN = 194599;
const int MAXQ = 1049659;
const long long MOD = 1802946053;
const int P = 1e9 + 7;
int n, q, indMp, val[MAXN];
int low[MAXAI + 2];
vector<int> primes[MAXN], idx[MAXN];
int dim[800];
int qryAns[MAXQ], l[MAXQ], r[MAXQ];
int maxLeft[15][MAXN];
void cit(int &x) {
  x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}
map<long long, int> cine;
int amLeft[8][MAXAI + 5];
void sita() {
  for (int i = 2; i <= MAXAI; ++i) {
    if (low[i] == 0) {
      low[i] = i;
      for (int j = i + i; j <= MAXAI; j += i)
        if (!low[j]) low[j] = i;
    }
  }
}
void decomp(int pos) {
  int x = val[pos];
  while (x > 1) {
    int coef = 0;
    int cf = low[x];
    while (x % cf == 0) coef ^= 1, x /= cf;
    if (coef == 1) primes[pos].push_back(cf);
  }
}
int nr[1 << 8];
void go(int pos) {
  int sz = primes[pos].size();
  for (int c = 0; c < 1 << sz; ++c) {
    nr[c] = 1;
    for (int b = 0; b < sz; ++b)
      if (!((1 << b) & c)) nr[c] *= primes[pos][b];
  }
  for (int c = 0; c < 1 << sz; ++c) {
    int alreadyD = dim[c];
    int tmp = nr[c];
    for (int oD = 0; oD <= 7; ++oD)
      maxLeft[oD + alreadyD][pos] =
          max(maxLeft[oD + alreadyD][pos], amLeft[oD][tmp]);
  }
  for (int c = 0; c < 1 << sz; ++c) {
    int alreadyD = dim[c], tmp = nr[c];
    amLeft[alreadyD][tmp] = pos;
  }
}
int aint[4 * MAXN + 5];
void update(int poz, int value, int st = 1, int dr = n, int node = 1) {
  if (st == dr) {
    aint[node] = value;
    return;
  }
  int mid = (st + dr) >> 1;
  if (poz <= mid)
    update(poz, value, st, mid, node * 2);
  else
    update(poz, value, mid + 1, dr, node * 2 + 1);
  aint[node] = max(aint[node * 2], aint[node * 2 + 1]);
}
int query(int x, int y, int st = 1, int dr = n, int node = 1) {
  if (x > dr || st > y) {
    return 0;
  }
  if (st >= x && dr <= y) return aint[node];
  int mid = (st + dr) >> 1;
  return max(query(x, y, st, mid, node * 2),
             query(x, y, mid + 1, dr, node * 2 + 1));
}
int first[MAXN];
int main() {
  sita();
  cit(n), cit(q);
  for (int i = 1; i <= 600; ++i)
    for (int j = 0; j < 10; ++j)
      if (i & (1 << j)) ++dim[i];
  for (int i = 1; i <= n; ++i) {
    cit(val[i]);
    decomp(i);
  }
  for (int i = 1; i <= q; ++i) cit(l[i]), cit(r[i]), qryAns[i] = 30;
  for (int i = 1; i <= n; ++i) {
    go(i);
  }
  for (int d = 14; d >= 0; --d) {
    for (int i = 1; i <= n; ++i) first[i] = n + 1;
    for (int i = 1; i <= n; ++i)
      first[maxLeft[d][i]] = min(first[maxLeft[d][i]], i);
    for (int i = n - 1; i >= 1; --i) first[i] = min(first[i], first[i + 1]);
    for (int i = 1; i <= q; ++i) {
      if (first[l[i]] <= r[i]) qryAns[i] = d;
    }
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", qryAns[i]);
}