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 <class T>
inline void gmin(T &x, const T &y) {
  x = x > y ? y : x;
}
template <class T>
inline void gmax(T &x, const T &y) {
  x = x < y ? y : x;
}
template <class T>
inline bool Gmin(T &x, const T &y) {
  return x > y ? x = y, 1 : 0;
}
template <class T>
inline bool Gmax(T &x, const T &y) {
  return x < y ? x = y, 1 : 0;
}
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Bufferhead, *Buffertail;
bool Terminal;
inline char Getchar() {
  if (Bufferhead == Buffertail) {
    int l = fread(buffer, 1, BufferSize, stdin);
    if (!l) {
      Terminal = 1;
      return 0;
    }
    Buffertail = (Bufferhead = buffer) + l;
  }
  return *Bufferhead++;
}
template <class T>
inline bool read(T &x) {
  x = 0;
  char c = Getchar(), rev = 0;
  while (c < '0' || c > '9') {
    rev |= c == '-';
    c = Getchar();
    if (Terminal) return 0;
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = Getchar();
  if (c == '.') {
    c = Getchar();
    double t = 0.1;
    while (c >= '0' && c <= '9')
      x = x + (c - '0') * t, c = Getchar(), t = t / 10;
  }
  x = rev ? -x : x;
  return 1;
}
template <class T1, class T2>
inline bool read(T1 &x, T2 &y) {
  return read(x) & read(y);
}
template <class T1, class T2, class T3>
inline bool read(T1 &x, T2 &y, T3 &z) {
  return read(x) & read(y) & read(z);
}
template <class T1, class T2, class T3, class T4>
inline bool read(T1 &x, T2 &y, T3 &z, T4 &w) {
  return read(x) & read(y) & read(z) & read(w);
}
inline bool reads(char *x) {
  char c = Getchar();
  while (c < 33 || c > 126) {
    c = Getchar();
    if (Terminal) return 0;
  }
  while (c >= 33 && c <= 126) (*x++) = c, c = Getchar();
  *x = 0;
  return 1;
}
template <class T>
inline void print(T x, const char c = '\n') {
  if (!x) {
    putchar('0');
    putchar(c);
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  int m = 0, a[20];
  while (x) a[m++] = x % 10, x /= 10;
  while (m--) putchar(a[m] + '0');
  putchar(c);
}
const int inf = 0x3f3f3f3f;
const int N = 200005, M = 5100000, mod = 1e9 + 7;
template <class T, class S>
inline void ch(T &x, const S y) {
  x = (x + y) % mod;
}
inline int exp(int x, int y, const int mod = ::mod) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = (long long)ans * x % mod;
    x = (long long)x * x % mod;
    y >>= 1;
  }
  return ans;
}
int n, q, a[N], trans[M + 5], cnt[M + 5];
int f[N][18], ans[M];
vector<int> pos[M + 5];
vector<pair<int, int> > Q[N];
inline void prepare() {
  static bool vis[M + 5];
  static int p[M];
  int n = M, tot = 0;
  trans[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) {
      p[++tot] = i;
      trans[i] = i;
      cnt[i] = 1;
    }
    for (int j = 1; j <= tot && p[j] * i <= n; j++) {
      vis[p[j] * i] = 1;
      if (i % p[j] == 0) {
        if (trans[i] % p[j] == 0)
          trans[p[j] * i] = trans[i] / p[j], cnt[p[j] * i] = cnt[i] - 1;
        else
          trans[p[j] * i] = trans[i] * p[j], cnt[p[j] * i] = cnt[i] + 1;
        break;
      }
      trans[p[j] * i] = trans[i] * p[j];
      cnt[p[j] * i] = cnt[i] + 1;
    }
  }
}
int main() {
  prepare();
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), a[i] = trans[a[i]], pos[a[i]].push_back(i);
  for (int i = 1; i <= M; i++) {
    if (trans[i] != i) continue;
    int cb = cnt[i] * 2;
    vector<int> b(0);
    for (int j = i; j <= M; j += i)
      for (int k : pos[j]) b.push_back(k);
    sort(b.begin(), b.end());
    static pair<int, int> s[N];
    int top = 0;
    if (!((int)b.size())) continue;
    for (int j : b) {
      for (int k = 1; k <= top; k++)
        gmax(f[j][cnt[a[j]] + s[k].first - cb], s[k].second);
      while (top && s[top].first >= cnt[a[j]]) top--;
      s[++top] = make_pair(cnt[a[j]], j);
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= 16; j++) {
      if (j) gmax(f[i][j], f[i][j - 1]);
      gmax(f[i][j], f[i - 1][j]);
    }
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    for (int j = 0; j <= 16; j++)
      if (f[r][j] >= l) {
        printf("%d\n", j);
        break;
      }
  }
  return 0;
}