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>
const int M = 5032107;
int prime[M + 1], top, f[M + 1], down[M + 1], n, q, a[200001], mx[M + 1];
int c[200001], ans[2000001];
void update(int ind, int num) {
  for (; ind; ind -= ind & -ind) c[ind] = std::min(c[ind], num);
}
int query(int ind) {
  int tot = 0x7f7f7f7f;
  for (; ind <= n; ind += ind & -ind) tot = std::min(tot, c[ind]);
  return tot;
}
std::vector<std::pair<int, int> > que[200001];
std::vector<int> Q[M + 1];
int stk[21];
bool pri[M + 1];
void add(int d, int i) {
  while (!Q[d].empty()) {
    update(Q[d].back(), f[a[Q[d].back()]] + f[a[i]] - (f[d] << 1));
    if (f[a[Q[d].back()]] > f[a[i]])
      Q[d].pop_back();
    else
      break;
  }
  Q[d].push_back(i);
}
void dfs(int x, int i, int prod) {
  if (x > top) {
    add(prod, i);
    return;
  }
  dfs(x + 1, i, prod);
  dfs(x + 1, i, prod * stk[x]);
}
int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
void write(int x) {
  if (x > 9) write(x / 10);
  putchar((x % 10) + '0');
}
int main() {
  down[1] = 1;
  mx[1] = 0;
  for (int i = 2; i <= M; i++) {
    if (!pri[i]) prime[++top] = i, f[i] = 1, down[i] = i, mx[i] = i;
    for (int j = 1; j <= top && i * prime[j] <= M; j++) {
      pri[i * prime[j]] = 1;
      mx[i * prime[j]] = std::max(mx[i], prime[j]);
      if (i % prime[j] == 0) {
        down[i * prime[j]] = down[i / prime[j]];
        break;
      }
      down[i * prime[j]] = down[i] * prime[j];
      f[i * prime[j]] = f[i] + 1;
    }
  }
  memset(c, 0x7f7f7f7f, sizeof c);
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) a[i] = down[read()];
  for (int i = 1, l, r; i <= q; i++)
    l = read(), r = read(), que[r].push_back(std::make_pair(l, i));
  for (int i = 1; i <= n; i++) {
    top = 0;
    for (int tem = a[i]; tem > 1; tem /= mx[tem]) {
      stk[++top] = mx[tem];
    }
    dfs(1, i, 1);
    for (int j = 0; j < que[i].size(); j++)
      ans[que[i][j].second] = query(que[i][j].first);
  }
  for (int i = 1; i <= q; i++) write(ans[i]), putchar('\n');
}