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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
vector<int> ps[200000], qs[200000];
int a[200000], l[1100000], r[1100000], ans[1100000];
int dp[5100000][10], D[33];
void fmain(int ID) {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    scanf("%d", a + i);
    int x = a[i];
    for (int j = 2; j * j <= x; j++)
      if (x % j == 0) {
        int cc = 0;
        while (x % j == 0) {
          cc++;
          x /= j;
        }
        if (cc & 1) ps[i].push_back(j);
      }
    if (x > 1) ps[i].push_back(x);
  }
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    scanf("%d%d", l + i, r + i);
    qs[r[i]].push_back(i);
  }
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    int N = ps[i].size();
    for (int(j) = 0; (j) < (int)(1 << N); (j)++) {
      int x = 1;
      for (int(z) = 0; (z) < (int)(N); (z)++)
        if (j & (1 << z)) x *= ps[i][z];
      int cc = N - __builtin_popcount(j);
      for (int(z) = 0; (z) < (int)(10); (z)++) maxtt(D[z + cc], dp[x][z]);
    }
    for (int id : qs[i]) {
      while (D[ans[id]] < l[id]) {
        ans[id]++;
      }
    }
    for (int(j) = 0; (j) < (int)(1 << N); (j)++) {
      int x = 1;
      for (int(z) = 0; (z) < (int)(N); (z)++)
        if (j & (1 << z)) x *= ps[i][z];
      int cc = N - __builtin_popcount(j);
      dp[x][cc] = i;
    }
  }
  for (int(i) = 1; (i) <= (int)(m); (i)++) printf("%d\n", ans[i]);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}