#include <bits/stdc++.h>
using namespace std;
const int N = 4e5 + 10;
namespace Seg {
int f[N << 1];
inline void Init() {
  for (int i = 0; i < N << 1; ++i) f[i] = -1e9;
}
inline void Update(int x, int y) {
  for (int i = N + x; i; i >>= 1) f[i] = max(f[i], y);
}
inline int Ask(int x, int y) {
  int ans = -1e9;
  for (int l = N + x - 1, r = N + y + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) ans = max(ans, f[l ^ 1]);
    if (r & 1) ans = max(ans, f[r ^ 1]);
  }
  return ans;
}
}  // namespace Seg
int n;
pair<int, int> p[N];
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; ++i) {
    scanf("%d", &x);
    p[i] = pair<int, int>(i - x, x);
  }
  sort(p + 1, p + n + 1);
  Seg::Init();
  Seg::Update(0, 0);
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    if (p[i].first >= 0) {
      int f = Seg::Ask(0, p[i].second - 1) + 1;
      ans = max(ans, f);
      Seg::Update(p[i].second, f);
    }
  printf("%d\n", ans);
  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 int N = 4e5 + 10;
namespace Seg {
int f[N << 1];
inline void Init() {
  for (int i = 0; i < N << 1; ++i) f[i] = -1e9;
}
inline void Update(int x, int y) {
  for (int i = N + x; i; i >>= 1) f[i] = max(f[i], y);
}
inline int Ask(int x, int y) {
  int ans = -1e9;
  for (int l = N + x - 1, r = N + y + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) ans = max(ans, f[l ^ 1]);
    if (r & 1) ans = max(ans, f[r ^ 1]);
  }
  return ans;
}
}  // namespace Seg
int n;
pair<int, int> p[N];
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; ++i) {
    scanf("%d", &x);
    p[i] = pair<int, int>(i - x, x);
  }
  sort(p + 1, p + n + 1);
  Seg::Init();
  Seg::Update(0, 0);
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    if (p[i].first >= 0) {
      int f = Seg::Ask(0, p[i].second - 1) + 1;
      ans = max(ans, f);
      Seg::Update(p[i].second, f);
    }
  printf("%d\n", ans);
  return 0;
}

```