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 N = 194598 + 77, NN = 5032107 + 77, L = 8, inf = 2000000000 + 77;
int n, q, a[N], d[NN], dp[N][L * 2], T[NN][L], A[NN], Mn[N << 2], lz[N << 2];
vector<pair<int, int> > Q[N];
inline void Shift(int l, int r, int id) {
  int x = lz[id];
  lz[id] = inf;
  if (x == inf) return;
  Mn[id] = min(Mn[id], x);
  if (r - l > 1)
    lz[id << 1] = min(lz[id << 1], x),
             lz[id << 1 ^ 1] = min(lz[id << 1 ^ 1], x);
}
void Update(int ql, int qr, int x, int l = 1, int r = n + 1, int id = 1) {
  Shift(l, r, id);
  if (qr <= l || r <= ql) return;
  if (ql <= l && r <= qr) {
    lz[id] = x;
    Shift(l, r, id);
    return;
  }
  int mid = ((l + r) >> 1);
  Update(ql, qr, x, l, mid, id << 1);
  Update(ql, qr, x, mid, r, id << 1 ^ 1);
  Mn[id] = min(Mn[id << 1], Mn[id << 1 ^ 1]);
}
int Get(int ql, int qr, int l = 1, int r = n + 1, int id = 1) {
  Shift(l, r, id);
  if (qr <= l || r <= ql) return 20;
  if (ql <= l && r <= qr) return Mn[id];
  int mid = ((l + r) >> 1);
  return min(Get(ql, qr, l, mid, id << 1), Get(ql, qr, mid, r, id << 1 ^ 1));
}
int main() {
  fill(Mn, Mn + (N << 2), inf);
  fill(lz, lz + (N << 2), inf);
  for (int i = 2; i < NN; ++i) {
    if (d[i] > 0) continue;
    for (int j = i; j < NN; j += i) d[j] = i;
  }
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    int f = 1;
    while (a[i] > 1) {
      int p = d[a[i]];
      int cnt = 0;
      while (a[i] % p == 0) ++cnt, a[i] /= p;
      if (cnt % 2) f *= p;
    }
    a[i] = f;
  }
  for (int l, r, i = 1; i <= q; ++i)
    scanf("%d %d", &l, &r), Q[r].push_back(make_pair(l, i));
  for (int i = 1; i <= n; ++i) {
    int f = a[i];
    vector<int> V;
    while (f > 1) V.push_back(d[f]), f /= d[f];
    int sz = V.size();
    for (int msk = 0; msk < (1 << sz); ++msk) {
      int p = 1;
      for (int j = 0; j < sz; ++j)
        if (msk & (1 << j)) p *= V[j];
      for (int j = 0; j < L; ++j)
        dp[i][j + sz - __builtin_popcount(msk)] =
            max(dp[i][j + sz - __builtin_popcount(msk)], T[p][j]);
    }
    for (int msk = 0; msk < (1 << sz); ++msk) {
      int p = 1;
      for (int j = 0; j < sz; ++j)
        if (msk & (1 << j)) p *= V[j];
      T[p][sz - __builtin_popcount(msk)] = i;
    }
    for (int j = 0; j < 2 * L; ++j) Update(1, dp[i][j] + 1, j);
    for (auto x : Q[i]) A[x.second] = Get(x.first, i + 1);
  }
  for (int i = 1; i <= q; ++i) printf("%d\n", A[i]);
  return 0;
}