#include <bits/stdc++.h>
using namespace std;
pair<int, int> pr[233333];
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first)
    return a.second > b.second;
  else
    return a.first < b.first;
}
int f[233333], st[233333], top;
int main() {
  ios_base::sync_with_stdio(false);
  int n, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    pr[i] = make_pair(x, i - x);
  }
  sort(pr + 1, pr + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    if (pr[i].second < 0) continue;
    int l = 0, r = top;
    while (l < r) {
      int mid = (l + r + 1) / 2;
      if (pr[i].second >= pr[st[mid]].second)
        l = mid;
      else
        r = mid - 1;
    }
    f[i] = l + 1;
    if (f[i] > top or pr[i].second < pr[st[l + 1]].second) st[l + 1] = i;
    if (f[i] > top) top = f[i];
  }
  cout << top << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

Mr. Chanek gives you a sequence a indexed from 1 to n. Define f(a) as the number of indices where a_i = i. 

You can pick an element from the current sequence and remove it, then concatenate the remaining elements together. For example, if you remove the 3-rd element from the sequence [4, 2, 3, 1], the resulting sequence will be [4, 2, 1]. 

You want to remove some elements from a in order to maximize f(a), using zero or more operations. Find the largest possible f(a).

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the initial length of the sequence.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2 ⋅ 10^5) — the initial sequence a.

Output

Output an integer denoting the largest f(a) that can be obtained by doing zero or more operations.

Examples

Input


7
2 1 4 2 5 3 7


Output


3


Input


4
4 2 3 1


Output


2

Note

In the first example, f(A) = 3 by doing the following operations.

[2,1,4,2,5,3,7] → [2,1,2,5,3,7] → [1,2,5,3,7] → [1,2,5,3] → [1,2,3]

In the second example, f(A) = 2 and no additional operation is needed.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
pair<int, int> pr[233333];
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first)
    return a.second > b.second;
  else
    return a.first < b.first;
}
int f[233333], st[233333], top;
int main() {
  ios_base::sync_with_stdio(false);
  int n, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    pr[i] = make_pair(x, i - x);
  }
  sort(pr + 1, pr + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    if (pr[i].second < 0) continue;
    int l = 0, r = top;
    while (l < r) {
      int mid = (l + r + 1) / 2;
      if (pr[i].second >= pr[st[mid]].second)
        l = mid;
      else
        r = mid - 1;
    }
    f[i] = l + 1;
    if (f[i] > top or pr[i].second < pr[st[l + 1]].second) st[l + 1] = i;
    if (f[i] > top) top = f[i];
  }
  cout << top << endl;
  return 0;
}

```