#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, m, t, f[200500], sb;
void add(int x, int y) {
  for (; x <= 200000; x += (-x & x)) {
    f[x] = max(f[x], y);
  }
}
int get(int x, int y = 0) {
  for (; x; x -= (-x & x)) {
    y = max(y, f[x]);
  }
  return y;
}
set<int> s[400500];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &k);
    if (k > i) continue;
    s[i + 200000 - k].insert(k);
  }
  for (i = 1; i <= 400000; i++) {
    for (auto j : s[i]) {
      sb = get(j - 1) + 1;
      add(j, sb);
    }
  }
  cout << get(200000);
}

### Prompt

Your task is to create 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;
int i, j, k, n, m, t, f[200500], sb;
void add(int x, int y) {
  for (; x <= 200000; x += (-x & x)) {
    f[x] = max(f[x], y);
  }
}
int get(int x, int y = 0) {
  for (; x; x -= (-x & x)) {
    y = max(y, f[x]);
  }
  return y;
}
set<int> s[400500];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &k);
    if (k > i) continue;
    s[i + 200000 - k].insert(k);
  }
  for (i = 1; i <= 400000; i++) {
    for (auto j : s[i]) {
      sb = get(j - 1) + 1;
      add(j, sb);
    }
  }
  cout << get(200000);
}

```