#include <bits/stdc++.h>
const int N = 200010;
int stk[N];
std::pair<int, int> pr[N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int n;
  std::cin >> n;
  for (int i = 1; i <= n; ++i) {
    int ai;
    std::cin >> ai;
    pr[i] = std::make_pair(ai, -i);
  }
  std::sort(pr + 1, pr + n + 1);
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    int x = -(pr[i].first + pr[i].second);
    if (x < 0) continue;
    int pos = std::upper_bound(stk + 1, stk + ans + 1, x) - stk;
    stk[pos] = x;
    ans = std::max(ans, pos);
  }
  std::cout << ans << '\n';
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
const int N = 200010;
int stk[N];
std::pair<int, int> pr[N];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int n;
  std::cin >> n;
  for (int i = 1; i <= n; ++i) {
    int ai;
    std::cin >> ai;
    pr[i] = std::make_pair(ai, -i);
  }
  std::sort(pr + 1, pr + n + 1);
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    int x = -(pr[i].first + pr[i].second);
    if (x < 0) continue;
    int pos = std::upper_bound(stk + 1, stk + ans + 1, x) - stk;
    stk[pos] = x;
    ans = std::max(ans, pos);
  }
  std::cout << ans << '\n';
  return 0;
}

```