#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long maxN = 300005;
const long long oo = 1e18 + 7;
long long n, a[maxN];
long long x, y, z, k;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  vector<pair<long long, long long> > ele;
  for (long long i = 1; i <= n; ++i) {
    cin >> x;
    if (x <= i) {
      ele.push_back(pair<long long, long long>(i - x, x));
    }
  }
  sort(ele.begin(), ele.end());
  vector<long long> lis;
  lis.push_back(-oo);
  long long bonus = 0;
  for (pair<long long, long long> ccc : ele) {
    long long cc = ccc.second;
    if (cc > lis.back())
      lis.push_back(cc);
    else {
      x = lower_bound(lis.begin(), lis.end(), cc) - lis.begin();
      lis[x] = min(lis[x], cc);
    }
  }
  cout << lis.size() - 1 << "\n";
}

### Prompt

Please formulate 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 long long mod = 1e9 + 7;
const long long maxN = 300005;
const long long oo = 1e18 + 7;
long long n, a[maxN];
long long x, y, z, k;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  vector<pair<long long, long long> > ele;
  for (long long i = 1; i <= n; ++i) {
    cin >> x;
    if (x <= i) {
      ele.push_back(pair<long long, long long>(i - x, x));
    }
  }
  sort(ele.begin(), ele.end());
  vector<long long> lis;
  lis.push_back(-oo);
  long long bonus = 0;
  for (pair<long long, long long> ccc : ele) {
    long long cc = ccc.second;
    if (cc > lis.back())
      lis.push_back(cc);
    else {
      x = lower_bound(lis.begin(), lis.end(), cc) - lis.begin();
      lis[x] = min(lis[x], cc);
    }
  }
  cout << lis.size() - 1 << "\n";
}

```