#include <bits/stdc++.h>
using namespace std;
struct node {
  int a, c;
} a[200005];
int n, dp[200005], maxlen;
bool cmp(node a, node b) { return a.a < b.a || a.a == b.a && a.c > b.c; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].a);
  for (int i = 1; i <= n; i++) a[i].c = i - a[i].a;
  sort(a + 1, a + 1 + n, cmp);
  dp[0] = -1e9;
  for (int i = 1; i <= n; i++)
    if (a[i].c >= 0) {
      if (dp[maxlen] <= a[i].c) {
        dp[++maxlen] = a[i].c;
      } else {
        int k = upper_bound(dp, dp + 1 + maxlen, a[i].c) - dp;
        dp[k] = a[i].c;
      }
    }
  printf("%d\n", maxlen);
  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;
struct node {
  int a, c;
} a[200005];
int n, dp[200005], maxlen;
bool cmp(node a, node b) { return a.a < b.a || a.a == b.a && a.c > b.c; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].a);
  for (int i = 1; i <= n; i++) a[i].c = i - a[i].a;
  sort(a + 1, a + 1 + n, cmp);
  dp[0] = -1e9;
  for (int i = 1; i <= n; i++)
    if (a[i].c >= 0) {
      if (dp[maxlen] <= a[i].c) {
        dp[++maxlen] = a[i].c;
      } else {
        int k = upper_bound(dp, dp + 1 + maxlen, a[i].c) - dp;
        dp[k] = a[i].c;
      }
    }
  printf("%d\n", maxlen);
  return 0;
}

```