#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-12;
const int INF = 1e9 + 10;
template <typename T>
using MinPQ = priority_queue<T, vector<T>, greater<T>>;
int n;
vector<int> a;
vector<pair<int, int>> pos;
void Solve() {
  scanf("%d", &n);
  a.resize(n + 1);
  pos.resize(n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pos[i - 1] = {a[i], i};
  }
  sort((pos).begin(), (pos).end(),
       [&](const pair<int, int>& x, const pair<int, int>& y) -> bool {
         if (x.first == y.first) return x.second > y.second;
         return x.first < y.first;
       });
  int ans = 0;
  map<int, int> f;
  f[0] = 0;
  for (auto [i, p] : pos) {
    int d = p - i;
    if (d < 0) continue;
    int w;
    {
      auto it = prev(f.upper_bound(d));
      w = it->second + 1;
      42;
    }
    ans = max(ans, w);
    f[d] = w;
    auto it = f.upper_bound(d);
    while (it != f.end() && it->second <= w) it = f.erase(it);
  }
  printf("%d\n", ans);
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; ++i) {
    Solve();
  }
  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;
const double EPS = 1e-12;
const int INF = 1e9 + 10;
template <typename T>
using MinPQ = priority_queue<T, vector<T>, greater<T>>;
int n;
vector<int> a;
vector<pair<int, int>> pos;
void Solve() {
  scanf("%d", &n);
  a.resize(n + 1);
  pos.resize(n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    pos[i - 1] = {a[i], i};
  }
  sort((pos).begin(), (pos).end(),
       [&](const pair<int, int>& x, const pair<int, int>& y) -> bool {
         if (x.first == y.first) return x.second > y.second;
         return x.first < y.first;
       });
  int ans = 0;
  map<int, int> f;
  f[0] = 0;
  for (auto [i, p] : pos) {
    int d = p - i;
    if (d < 0) continue;
    int w;
    {
      auto it = prev(f.upper_bound(d));
      w = it->second + 1;
      42;
    }
    ans = max(ans, w);
    f[d] = w;
    auto it = f.upper_bound(d);
    while (it != f.end() && it->second <= w) it = f.erase(it);
  }
  printf("%d\n", ans);
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; ++i) {
    Solve();
  }
  return 0;
}

```