#include <bits/stdc++.h>
using namespace std;
long long ans, n, x;
vector<int> dp;
vector<pair<int, int>> A;
int main() {
  cin >> n;
  int curr = 0;
  dp.assign(n + 1, 0);
  A.resize(n + 1);
  A[0] = {-1000000, 0};
  for (int i = 1; i <= n; i++) {
    cin >> x;
    A[i] = {i - x, x};
  }
  sort(A.begin(), A.end());
  for (int i = 1; i <= n; i++) {
    if (A[i].first < 0)
      continue;
    else {
      int val = A[i].second;
      if (val > dp[curr]) {
        curr++;
        dp[curr] = val;
      } else {
        int l = 0, r = curr, mid;
        while (r - l > 1) {
          mid = (l + r) / 2;
          if (val <= dp[mid])
            r = mid;
          else
            l = mid;
        }
        dp[r] = val;
      }
    }
  }
  cout << curr << endl;
}

### 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;
long long ans, n, x;
vector<int> dp;
vector<pair<int, int>> A;
int main() {
  cin >> n;
  int curr = 0;
  dp.assign(n + 1, 0);
  A.resize(n + 1);
  A[0] = {-1000000, 0};
  for (int i = 1; i <= n; i++) {
    cin >> x;
    A[i] = {i - x, x};
  }
  sort(A.begin(), A.end());
  for (int i = 1; i <= n; i++) {
    if (A[i].first < 0)
      continue;
    else {
      int val = A[i].second;
      if (val > dp[curr]) {
        curr++;
        dp[curr] = val;
      } else {
        int l = 0, r = curr, mid;
        while (r - l > 1) {
          mid = (l + r) / 2;
          if (val <= dp[mid])
            r = mid;
          else
            l = mid;
        }
        dp[r] = val;
      }
    }
  }
  cout << curr << endl;
}

```