#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int n, a[200200];
set<pair<int, int>> s[200200];
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<pair<int, int>> v;
  for (int i = 1; i <= n; i++)
    if (a[i] <= i) v.push_back({i - a[i], a[i]});
  sort(v.begin(), v.end());
  for (auto [x, y] : v) {
    int l = 1, r = n;
    while (l <= r) {
      int m = (l + r) / 2;
      auto it = s[m].lower_bound({x, y});
      if (it != s[m].begin() && prev(it)->second < y)
        l = m + 1;
      else
        r = m - 1;
    }
    auto it = next(s[l].insert({x, y}).first);
    while (it != s[l].end() && it->second >= y) it = s[l].erase(it);
  }
  int l = 1;
  while (!s[l].empty()) l++;
  cout << l - 1;
}

### 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;
using ll = long long;
int n, a[200200];
set<pair<int, int>> s[200200];
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  vector<pair<int, int>> v;
  for (int i = 1; i <= n; i++)
    if (a[i] <= i) v.push_back({i - a[i], a[i]});
  sort(v.begin(), v.end());
  for (auto [x, y] : v) {
    int l = 1, r = n;
    while (l <= r) {
      int m = (l + r) / 2;
      auto it = s[m].lower_bound({x, y});
      if (it != s[m].begin() && prev(it)->second < y)
        l = m + 1;
      else
        r = m - 1;
    }
    auto it = next(s[l].insert({x, y}).first);
    while (it != s[l].end() && it->second >= y) it = s[l].erase(it);
  }
  int l = 1;
  while (!s[l].empty()) l++;
  cout << l - 1;
}

```