#include <bits/stdc++.h>
using namespace std;
long long LIS(vector<long long>& v) {
  long long n = v.size();
  if (!n) return 0;
  vector<long long> dp;
  dp = {v[0]};
  for (long long i = 1; i < n; ++i) {
    auto it = lower_bound(dp.begin(), dp.end(), v[i]);
    if (it == dp.end())
      dp.push_back(v[i]);
    else
      *it = v[i];
  }
  return dp.size();
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a;
  for (long long i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    if (i >= x) {
      a.push_back({i - x, x});
    }
  }
  sort(a.begin(), a.end());
  vector<long long> b;
  for (auto& [i, j] : a) {
    b.push_back(j);
  }
  cout << LIS(b);
  return 0;
}

### Prompt

Generate 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;
long long LIS(vector<long long>& v) {
  long long n = v.size();
  if (!n) return 0;
  vector<long long> dp;
  dp = {v[0]};
  for (long long i = 1; i < n; ++i) {
    auto it = lower_bound(dp.begin(), dp.end(), v[i]);
    if (it == dp.end())
      dp.push_back(v[i]);
    else
      *it = v[i];
  }
  return dp.size();
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<pair<long long, long long>> a;
  for (long long i = 1; i <= n; ++i) {
    long long x;
    cin >> x;
    if (i >= x) {
      a.push_back({i - x, x});
    }
  }
  sort(a.begin(), a.end());
  vector<long long> b;
  for (auto& [i, j] : a) {
    b.push_back(j);
  }
  cout << LIS(b);
  return 0;
}

```