#include <bits/stdc++.h>
using namespace std;
const int maxn = 2 * 1e5 + 10, maxm = 2 * 8 * maxn, mod = 998244353;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1), eps = 1e-7;
int n;
int a[maxn];
int f[maxn];
int tr[maxn];
int lowbit(int x) { return x & (-x); }
void modify(int x, int k) {
  while (x <= n) {
    tr[x] = max(tr[x], k);
    x += lowbit(x);
  }
}
int query(int x) {
  int res = 0;
  while (x) {
    res = max(res, tr[x]);
    x -= lowbit(x);
  }
  return res;
}
pair<int, int> b[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    b[i] = {i - a[i], a[i]};
  }
  sort(b + 1, b + 1 + n);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (b[i].first >= 0) {
      f[i] = query(b[i].second) + 1;
      modify(b[i].second + 1, f[i]);
      ans = max(ans, f[i]);
    }
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please create a solution in CPP 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 = 2 * 1e5 + 10, maxm = 2 * 8 * maxn, mod = 998244353;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1), eps = 1e-7;
int n;
int a[maxn];
int f[maxn];
int tr[maxn];
int lowbit(int x) { return x & (-x); }
void modify(int x, int k) {
  while (x <= n) {
    tr[x] = max(tr[x], k);
    x += lowbit(x);
  }
}
int query(int x) {
  int res = 0;
  while (x) {
    res = max(res, tr[x]);
    x -= lowbit(x);
  }
  return res;
}
pair<int, int> b[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    b[i] = {i - a[i], a[i]};
  }
  sort(b + 1, b + 1 + n);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (b[i].first >= 0) {
      f[i] = query(b[i].second) + 1;
      modify(b[i].second + 1, f[i]);
      ans = max(ans, f[i]);
    }
  }
  cout << ans << "\n";
  return 0;
}

```