#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int t[N * 4];
struct dt {
  int x, id;
  bool operator<(const dt& a) const {
    return id - x != a.id - a.x ? id - x < a.id - a.x : x < a.x;
  }
} f[N];
int n, m;
inline int ask(int x, int l, int r, int L, int R) {
  if (l == L && r == R) return t[x];
  int mid = (l + r) / 2;
  if (R <= mid) return ask(x * 2, l, mid, L, R);
  if (L > mid) return ask(x * 2 + 1, mid + 1, r, L, R);
  return max(ask(x * 2, l, mid, L, mid),
             ask(x * 2 + 1, mid + 1, r, mid + 1, R));
}
inline void modify(int x, int l, int r, int p, int y) {
  t[x] = max(t[x], y);
  if (l == r) return;
  int mid = (l + r) / 2;
  if (p <= mid)
    modify(x * 2, l, mid, p, y);
  else
    modify(x * 2 + 1, mid + 1, r, p, y);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; ++i) {
    scanf("%d", &x);
    if (x <= i) f[++m] = {x, i};
  }
  sort(f + 1, f + 1 + m);
  for (int i = 1; i <= m; ++i) {
    int p = (f[i].x == 1 ? 0 : ask(1, 1, n, 1, f[i].x - 1));
    modify(1, 1, n, f[i].x, p + 1);
  }
  cout << t[1] << endl;
  return 0;
}

### Prompt

Generate 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 N = 200005;
int t[N * 4];
struct dt {
  int x, id;
  bool operator<(const dt& a) const {
    return id - x != a.id - a.x ? id - x < a.id - a.x : x < a.x;
  }
} f[N];
int n, m;
inline int ask(int x, int l, int r, int L, int R) {
  if (l == L && r == R) return t[x];
  int mid = (l + r) / 2;
  if (R <= mid) return ask(x * 2, l, mid, L, R);
  if (L > mid) return ask(x * 2 + 1, mid + 1, r, L, R);
  return max(ask(x * 2, l, mid, L, mid),
             ask(x * 2 + 1, mid + 1, r, mid + 1, R));
}
inline void modify(int x, int l, int r, int p, int y) {
  t[x] = max(t[x], y);
  if (l == r) return;
  int mid = (l + r) / 2;
  if (p <= mid)
    modify(x * 2, l, mid, p, y);
  else
    modify(x * 2 + 1, mid + 1, r, p, y);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; ++i) {
    scanf("%d", &x);
    if (x <= i) f[++m] = {x, i};
  }
  sort(f + 1, f + 1 + m);
  for (int i = 1; i <= m; ++i) {
    int p = (f[i].x == 1 ? 0 : ask(1, 1, n, 1, f[i].x - 1));
    modify(1, 1, n, f[i].x, p + 1);
  }
  cout << t[1] << endl;
  return 0;
}

```