#include <bits/stdc++.h>
using namespace std;
int a[500010];
pair<int, int> b[500010];
int c[500010];
int T, n, tot;
int main(void) {
  T = 1;
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      b[i] = {a[i], -i};
    }
    sort(b + 1, b + 1 + n);
    for (int i = 1; i <= n; ++i) {
      if (-b[i].second >= b[i].first) {
        int x = -(b[i].second + b[i].first);
        int pos = upper_bound(c + 1, c + 1 + tot, x) - c;
        c[pos] = x;
        tot = max(pos, tot);
      }
    }
    printf("%d\n", tot);
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int a[500010];
pair<int, int> b[500010];
int c[500010];
int T, n, tot;
int main(void) {
  T = 1;
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      b[i] = {a[i], -i};
    }
    sort(b + 1, b + 1 + n);
    for (int i = 1; i <= n; ++i) {
      if (-b[i].second >= b[i].first) {
        int x = -(b[i].second + b[i].first);
        int pos = upper_bound(c + 1, c + 1 + tot, x) - c;
        c[pos] = x;
        tot = max(pos, tot);
      }
    }
    printf("%d\n", tot);
  }
  return 0;
}

```