#include <bits/stdc++.h>
using namespace std;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int n, a[500005], f[500005];
;
pair<int, int> P[500005];
int get(int u) {
  int res = 0;
  for (int i = u; i > 0; i -= i & -i) {
    res = max(res, f[i]);
  }
  return res;
}
void update(int u, int gt) {
  for (int i = u; i < 500005; i += i & -i) f[i] = max(f[i], gt);
}
void solve() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    cin >> a[i];
    P[i] = make_pair(i + n - a[i], i);
  }
  sort(P + 1, P + n + 1);
  memset(f, 0, sizeof(f));
  int res = 0;
  update(1, 0);
  for (int i = (1); i <= (n); ++i) {
    int id = P[i].second;
    if (P[i].first < n) continue;
    int u = a[id];
    int A = get(u) + 1;
    res = max(res, A);
    update(u + 1, A);
  }
  cout << res << endl;
}
int main() {
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int n, a[500005], f[500005];
;
pair<int, int> P[500005];
int get(int u) {
  int res = 0;
  for (int i = u; i > 0; i -= i & -i) {
    res = max(res, f[i]);
  }
  return res;
}
void update(int u, int gt) {
  for (int i = u; i < 500005; i += i & -i) f[i] = max(f[i], gt);
}
void solve() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = (1); i <= (n); ++i) {
    cin >> a[i];
    P[i] = make_pair(i + n - a[i], i);
  }
  sort(P + 1, P + n + 1);
  memset(f, 0, sizeof(f));
  int res = 0;
  update(1, 0);
  for (int i = (1); i <= (n); ++i) {
    int id = P[i].second;
    if (P[i].first < n) continue;
    int u = a[id];
    int A = get(u) + 1;
    res = max(res, A);
    update(u + 1, A);
  }
  cout << res << endl;
}
int main() {
  solve();
  return 0;
}

```