#include <bits/stdc++.h>
using namespace std;
inline int rd() {
  int x = 0;
  char ch, t = 0;
  while (!isdigit(ch = getchar())) t |= ch == '-';
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return x = t ? -x : x;
}
inline long long maxll(long long x, long long y) { return x > y ? x : y; }
inline long long minll(long long x, long long y) { return x < y ? x : y; }
inline long long absll(long long x) { return x > 0ll ? x : -x; }
inline long long gcd(long long x, long long y) {
  return (y == 0) ? x : gcd(y, x % y);
}
int n, ans;
int minn[200005];
pair<int, int> a[200005];
int main() {
  n = rd();
  for (int i = 1; i <= n; i++) a[i] = pair<int, int>(rd(), -i);
  sort(a + 1, a + n + 1);
  for (int i = 1, pos, x; i <= n; i++) {
    x = -a[i].second - a[i].first;
    if (x < 0) continue;
    pos = upper_bound(minn + 1, minn + ans + 1, x) - minn;
    minn[pos] = x, ans = max(ans, pos);
  }
  printf("%d\n", ans);
  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;
inline int rd() {
  int x = 0;
  char ch, t = 0;
  while (!isdigit(ch = getchar())) t |= ch == '-';
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return x = t ? -x : x;
}
inline long long maxll(long long x, long long y) { return x > y ? x : y; }
inline long long minll(long long x, long long y) { return x < y ? x : y; }
inline long long absll(long long x) { return x > 0ll ? x : -x; }
inline long long gcd(long long x, long long y) {
  return (y == 0) ? x : gcd(y, x % y);
}
int n, ans;
int minn[200005];
pair<int, int> a[200005];
int main() {
  n = rd();
  for (int i = 1; i <= n; i++) a[i] = pair<int, int>(rd(), -i);
  sort(a + 1, a + n + 1);
  for (int i = 1, pos, x; i <= n; i++) {
    x = -a[i].second - a[i].first;
    if (x < 0) continue;
    pos = upper_bound(minn + 1, minn + ans + 1, x) - minn;
    minn[pos] = x, ans = max(ans, pos);
  }
  printf("%d\n", ans);
  return 0;
}

```