#include <bits/stdc++.h>
using namespace std;
const long long Mod = 1000000007LL;
const int N = 2e5 + 10;
const long long Inf = 2242545357980376863LL;
const long long Log = 30;
int fen[N];
void Add(int idx, int x) {
  idx++;
  for (; idx < N; idx += (idx & (-idx))) fen[idx] = max(fen[idx], x);
}
int Get(int idx) {
  int res = -N;
  idx++;
  for (; idx; idx -= (idx & (-idx))) res = max(fen[idx], res);
  return res;
}
vector<int> V[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int v;
  for (int i = 1; i <= n; i++) {
    cin >> v;
    V[v].push_back(i);
  }
  fill(fen, fen + N, -N);
  Add(0, 0);
  int ans = 0;
  for (int i = 1; i < N; i++) {
    reverse(V[i].begin(), V[i].end());
    for (auto idx : V[i]) {
      if (idx < i) continue;
      int res = Get(idx - i) + 1;
      ans = max(ans, res);
      Add(idx - i, res);
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 1000000007LL;
const int N = 2e5 + 10;
const long long Inf = 2242545357980376863LL;
const long long Log = 30;
int fen[N];
void Add(int idx, int x) {
  idx++;
  for (; idx < N; idx += (idx & (-idx))) fen[idx] = max(fen[idx], x);
}
int Get(int idx) {
  int res = -N;
  idx++;
  for (; idx; idx -= (idx & (-idx))) res = max(fen[idx], res);
  return res;
}
vector<int> V[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int v;
  for (int i = 1; i <= n; i++) {
    cin >> v;
    V[v].push_back(i);
  }
  fill(fen, fen + N, -N);
  Add(0, 0);
  int ans = 0;
  for (int i = 1; i < N; i++) {
    reverse(V[i].begin(), V[i].end());
    for (auto idx : V[i]) {
      if (idx < i) continue;
      int res = Get(idx - i) + 1;
      ans = max(ans, res);
      Add(idx - i, res);
    }
  }
  cout << ans << '\n';
  return 0;
}

```