#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, tr[N];
int lowbit(int x) { return x & (-x); }
void update(int x, int t) {
  for (int i = x; i <= n; i += lowbit(i)) tr[i] = max(tr[i], t);
}
int ask(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= lowbit(i)) ans = max(ans, tr[i]);
  return ans;
}
set<int> e[N];
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    int j = i - x;
    if (j >= 0 && j <= i - 1) e[j].insert(x);
  }
  for (int i = 0; i <= n; i++) {
    for (auto x : e[i]) {
      int j = ask(x - 1);
      update(x, j + 1);
    }
  }
  printf("%d", ask(n));
  return 0;
}

### 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;
const int N = 2e5 + 5;
int n, tr[N];
int lowbit(int x) { return x & (-x); }
void update(int x, int t) {
  for (int i = x; i <= n; i += lowbit(i)) tr[i] = max(tr[i], t);
}
int ask(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= lowbit(i)) ans = max(ans, tr[i]);
  return ans;
}
set<int> e[N];
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    int j = i - x;
    if (j >= 0 && j <= i - 1) e[j].insert(x);
  }
  for (int i = 0; i <= n; i++) {
    for (auto x : e[i]) {
      int j = ask(x - 1);
      update(x, j + 1);
    }
  }
  printf("%d", ask(n));
  return 0;
}

```