#include <bits/stdc++.h>
using namespace std;
int fact[100005];
bitset<100005> p;
int cnt[100005];
int best[100005];
int arr[100005];
int bestFactor(int n) {
  if (fact[n] == 0) {
    return n;
  }
  int curFactor = fact[n];
  int bestChildFactor = bestFactor(n / curFactor);
  if (best[curFactor] == -1) {
    return bestChildFactor;
  }
  if (best[bestChildFactor] == -1) {
    return curFactor;
  }
  if (cnt[best[curFactor]] > cnt[best[bestChildFactor]]) {
    return curFactor;
  }
  return bestChildFactor;
}
void updateFactors(int n, int idx) {
  if (fact[n] == 0) {
    best[n] = idx;
    return;
  }
  best[fact[n]] = idx;
  return updateFactors(n / fact[n], idx);
}
int main() {
  memset(best, -1, sizeof best);
  p.set();
  p[0] = p[1] = false;
  for (int i = 2, to = sqrt(100005) + 1; i <= to; i++) {
    if (p[i]) {
      for (int j = i * i; j < 100005; j += i) {
        if (fact[j] == 0) fact[j] = i;
        p[j] = false;
      }
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    cnt[i] = 1;
  }
  updateFactors(arr[0], 0);
  int maxi = 1;
  for (int i = 1; i < n; i++) {
    int bestFact = bestFactor(arr[i]);
    if (best[bestFact] != -1) {
      cnt[i] = cnt[best[bestFact]] + 1;
      maxi = max(maxi, cnt[i]);
    }
    updateFactors(arr[i], i);
  }
  cout << maxi;
  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;
int fact[100005];
bitset<100005> p;
int cnt[100005];
int best[100005];
int arr[100005];
int bestFactor(int n) {
  if (fact[n] == 0) {
    return n;
  }
  int curFactor = fact[n];
  int bestChildFactor = bestFactor(n / curFactor);
  if (best[curFactor] == -1) {
    return bestChildFactor;
  }
  if (best[bestChildFactor] == -1) {
    return curFactor;
  }
  if (cnt[best[curFactor]] > cnt[best[bestChildFactor]]) {
    return curFactor;
  }
  return bestChildFactor;
}
void updateFactors(int n, int idx) {
  if (fact[n] == 0) {
    best[n] = idx;
    return;
  }
  best[fact[n]] = idx;
  return updateFactors(n / fact[n], idx);
}
int main() {
  memset(best, -1, sizeof best);
  p.set();
  p[0] = p[1] = false;
  for (int i = 2, to = sqrt(100005) + 1; i <= to; i++) {
    if (p[i]) {
      for (int j = i * i; j < 100005; j += i) {
        if (fact[j] == 0) fact[j] = i;
        p[j] = false;
      }
    }
  }
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    cnt[i] = 1;
  }
  updateFactors(arr[0], 0);
  int maxi = 1;
  for (int i = 1; i < n; i++) {
    int bestFact = bestFactor(arr[i]);
    if (best[bestFact] != -1) {
      cnt[i] = cnt[best[bestFact]] + 1;
      maxi = max(maxi, cnt[i]);
    }
    updateFactors(arr[i], i);
  }
  cout << maxi;
  return 0;
}

```