#include <bits/stdc++.h>
using namespace std;
const int N = 100005, inf = 100000000;
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int a[N], dp[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  dp[0] = 0;
  for (int i = 1; i <= n; ++i) {
    int& res = dp[i];
    res = 1;
    for (int j = i - 1; j >= 1 && i - j <= 100; --j) {
      if (gcd(a[i - 1], a[j - 1]) > 1) {
        res = max(res, 1 + dp[j]);
      }
    }
  }
  int res = 0;
  for (int i = 0; i <= n; ++i) {
    res = max(res, dp[i]);
  }
  printf("%d\n", res);
  return (0);
}

### Prompt

Generate a CPP solution to the following problem:

Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks n integers a1, a2, ..., an are good.

Now she is interested in good sequences. A sequence x1, x2, ..., xk is called good if it satisfies the following three conditions:

  * The sequence is strictly increasing, i.e. xi < xi + 1 for each i (1 ≤ i ≤ k - 1). 
  * No two adjacent elements are coprime, i.e. gcd(xi, xi + 1) > 1 for each i (1 ≤ i ≤ k - 1) (where gcd(p, q) denotes the greatest common divisor of the integers p and q). 
  * All elements of the sequence are good integers. 



Find the length of the longest good sequence.

Input

The input consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 105) — the number of good integers. The second line contains a single-space separated list of good integers a1, a2, ..., an in strictly increasing order (1 ≤ ai ≤ 105; ai < ai + 1).

Output

Print a single integer — the length of the longest good sequence.

Examples

Input

5
2 3 4 6 9


Output

4


Input

9
1 2 3 5 6 7 8 9 10


Output

4

Note

In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100005, inf = 100000000;
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
int a[N], dp[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  dp[0] = 0;
  for (int i = 1; i <= n; ++i) {
    int& res = dp[i];
    res = 1;
    for (int j = i - 1; j >= 1 && i - j <= 100; --j) {
      if (gcd(a[i - 1], a[j - 1]) > 1) {
        res = max(res, 1 + dp[j]);
      }
    }
  }
  int res = 0;
  for (int i = 0; i <= n; ++i) {
    res = max(res, dp[i]);
  }
  printf("%d\n", res);
  return (0);
}

```