#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
int n, a[MAXN];
pair<int, int> ta[MAXN];
int seq[MAXN];
void solve() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (n); ++i) {
    ta[i].first = i - a[i];
    ta[i].second = a[i];
  }
  sort(ta + 1, ta + 1 + n,
       [](const pair<int, int> &lhs, const pair<int, int> &rhs) {
         return lhs.first < rhs.first ||
                (lhs.first == rhs.first && lhs.second < rhs.second);
       });
  int i = 1;
  while (i <= n && ta[i].first < 0) ++i;
  if (i > n) {
    cout << "0\n";
    return;
  }
  int ans = 1;
  seq[1] = ta[i].second;
  for (++i; i <= n; ++i) {
    auto p = lower_bound(seq + 1, seq + 1 + ans, ta[i].second);
    *p = ta[i].second;
    ans = max(ans, int(p - seq));
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _;
  _ = 1;
  while (_--) {
    solve();
  }
  return 0;
}

### 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 int MAXN = 2e5 + 10;
int n, a[MAXN];
pair<int, int> ta[MAXN];
int seq[MAXN];
void solve() {
  cin >> n;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  for (int i = (1); i <= (n); ++i) {
    ta[i].first = i - a[i];
    ta[i].second = a[i];
  }
  sort(ta + 1, ta + 1 + n,
       [](const pair<int, int> &lhs, const pair<int, int> &rhs) {
         return lhs.first < rhs.first ||
                (lhs.first == rhs.first && lhs.second < rhs.second);
       });
  int i = 1;
  while (i <= n && ta[i].first < 0) ++i;
  if (i > n) {
    cout << "0\n";
    return;
  }
  int ans = 1;
  seq[1] = ta[i].second;
  for (++i; i <= n; ++i) {
    auto p = lower_bound(seq + 1, seq + 1 + ans, ta[i].second);
    *p = ta[i].second;
    ans = max(ans, int(p - seq));
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int _;
  _ = 1;
  while (_--) {
    solve();
  }
  return 0;
}

```