#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
long long mod = 1e9 + 7;
long double eps = 1e-12;
long double pi = acosl(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ifstream in("input.txt");
ofstream out("output.txt");
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using uc = unsigned char;
using us = unsigned short;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int>> b;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] <= i + 1) b.push_back({i + 1 - a[i], a[i]});
    a[i] = i + 1 - a[i];
  }
  sort(b.begin(), b.end());
  if (b.size() == 0) {
    cout << 0;
    return;
  }
  vector<int> ans(1, b[0].second);
  for (int i = 1; i < b.size(); i++) {
    int id = lower_bound(ans.begin(), ans.end(), b[i].second) - ans.begin();
    if (id == ans.size()) {
      ans.push_back(b[i].second);
    } else {
      ans[id] = b[i].second;
    }
  }
  cout << ans.size();
}
int main() {
  cout << setprecision(20) << fixed;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  clock_t t1 = clock();
  int t = 1;
  int cs = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
long long mod = 1e9 + 7;
long double eps = 1e-12;
long double pi = acosl(-1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ifstream in("input.txt");
ofstream out("output.txt");
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using uc = unsigned char;
using us = unsigned short;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<pair<int, int>> b;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] <= i + 1) b.push_back({i + 1 - a[i], a[i]});
    a[i] = i + 1 - a[i];
  }
  sort(b.begin(), b.end());
  if (b.size() == 0) {
    cout << 0;
    return;
  }
  vector<int> ans(1, b[0].second);
  for (int i = 1; i < b.size(); i++) {
    int id = lower_bound(ans.begin(), ans.end(), b[i].second) - ans.begin();
    if (id == ans.size()) {
      ans.push_back(b[i].second);
    } else {
      ans[id] = b[i].second;
    }
  }
  cout << ans.size();
}
int main() {
  cout << setprecision(20) << fixed;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  clock_t t1 = clock();
  int t = 1;
  int cs = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```