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 + 7;
const int maxM = 5100000;
int a[maxN];
int pos[maxM][8];
int nex[maxM];
int L[maxN][15];
vector<int> vec;
struct tree {
  int MIN[maxN * 4];
  void bulidtree(int k, int l, int r, int p) {
    if (l == r) {
      MIN[k] = L[l][p];
      return;
    }
    int mid = (l + r) >> 1;
    bulidtree((k << 1), l, mid, p);
    bulidtree((k << 1 | 1), mid + 1, r, p);
    MIN[k] = min(MIN[(k << 1)], MIN[(k << 1 | 1)]);
  }
  int query(int k, int l, int r, int x, int y) {
    if (x <= l && r <= y) return MIN[k];
    int mid = (l + r) >> 1;
    int ret = maxN;
    if (x <= mid) ret = min(ret, query((k << 1), l, mid, x, y));
    if (y > mid) ret = min(ret, query((k << 1 | 1), mid + 1, r, x, y));
    return ret;
  }
} T[15];
int main() {
  for (int i = 2; i < maxM; i++) {
    if (nex[i]) continue;
    nex[i] = i;
    long long j = 1LL * i * i;
    while (j < maxM) {
      nex[j] = i;
      j += i;
    }
  }
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < 14; j++) L[i][j] = n + 10;
  for (int i = 1; i < maxM; i++)
    for (int j = 0; j <= 7; j++) pos[i][j] = n + 10;
  for (int i = n; i > 0; i--) {
    vec.clear();
    int x = a[i];
    while (x > 1) {
      int y = nex[x];
      int cnt = 0;
      while (x % y == 0) x /= y, cnt++;
      if (cnt & 1) vec.push_back(y);
    }
    for (int s = 0; s < (1 << vec.size()); s++) {
      int x = 1;
      int cnt = vec.size();
      for (int j = 0; j < vec.size(); j++)
        if (s & (1 << j)) x *= vec[j], cnt--;
      for (int k = 0; k <= 7; k++)
        L[i][cnt + k] = min(L[i][cnt + k], pos[x][k]);
      pos[x][cnt] = i;
    }
  }
  for (int i = 0; i <= 14; i++) T[i].bulidtree(1, 1, n, i);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    for (int i = 0; i <= 14; i++) {
      if (T[i].query(1, 1, n, x, y) <= y) {
        printf("%d\n", i);
        break;
      }
    }
  }
  return 0;
}