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 INF = 1 << 30;
template <typename T>
struct SparseTable {
  vector<vector<T> > st;
  SparseTable() {}
  SparseTable(const vector<T> &v) {
    int b = 0;
    while ((1 << b) <= v.size()) ++b;
    st.assign(b, vector<T>(1 << b));
    for (int i = 0; i < v.size(); i++) {
      st[0][i] = v[i];
    }
    for (int i = 1; i < b; i++) {
      for (int j = 0; j + (1 << i) <= (1 << b); j++) {
        st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
      }
    }
  }
  inline T rmq(int l, int r) {
    int b = 32 - __builtin_clz(r - l) - 1;
    return (min(st[b][l], st[b][r - (1 << b)]));
  }
};
int N, P, A[194598];
int d[5032109];
int latte[5032109][8];
int main() {
  fill_n(*latte, 5032109 * 8, INF);
  for (int i = 1; i <= 5032108; i++) d[i] = i;
  for (int i = 2; i * i <= 5032108; i++) {
    if (d[i] != i) continue;
    for (int j = i + i; j <= 5032108; j += i) d[j] = min(d[j], i);
  }
  scanf("%d %d", &N, &P);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
  }
  vector<vector<int> > dp(15, vector<int>(N, INF));
  for (int i = N - 1; i >= 0; i--) {
    set<int> st;
    while (A[i] != 1) {
      if (st.count(d[A[i]]))
        st.erase(d[A[i]]);
      else
        st.emplace(d[A[i]]);
      A[i] /= d[A[i]];
    }
    vector<int> uku;
    for (auto &p : st) uku.emplace_back(p);
    for (int j = 0; j < (1 << uku.size()); j++) {
      int mul = 1, add = 0;
      for (int k = 0; k < uku.size(); k++) {
        if ((j >> k) & 1)
          mul *= uku[k];
        else
          ++add;
      }
      for (int k = 0; k < 8; k++) {
        dp[k + add][i] = min(dp[k + add][i], latte[mul][k]);
      }
    }
    for (int j = 0; j < (1 << uku.size()); j++) {
      int add = 0, mul = 1;
      for (int k = 0; k < uku.size(); k++) {
        if ((j >> k) & 1)
          mul *= uku[k];
        else
          ++add;
      }
      latte[mul][add] = i;
    }
  }
  auto f = [&](int a, int b) { return min(a, b); };
  vector<SparseTable<int> > seg;
  for (int i = 0; i < 15; i++) {
    seg.emplace_back(SparseTable<int>(dp[i]));
  }
  for (int i = 0; i < P; i++) {
    int L, R;
    scanf("%d %d", &L, &R);
    --L;
    for (int j = 0; j < 15; j++) {
      if (seg[j].rmq(L, R) < R) {
        printf("%d\n", j);
        break;
      }
    }
  }
}