#include <bits/stdc++.h>
struct Node {
  int val;
  int idx;
};
const int maxn = 2e5 + 10;
int n;
Node arr[maxn];
int b[maxn], cnt;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i].idx);
    arr[i].val = i - arr[i].idx;
  }
  std::sort(arr + 1, arr + 1 + n, [](const Node &a, const Node &b) {
    if (a.val == b.val) return a.idx < b.idx;
    return a.val < b.val;
  });
  for (int i = 1; i <= n; i++) {
    if (arr[i].val < 0) continue;
    int dex = std::lower_bound(b + 1, b + cnt + 1, arr[i].idx) - b;
    b[dex] = arr[i].idx;
    cnt = std::max(cnt, dex);
  }
  printf("%d\n", cnt);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
struct Node {
  int val;
  int idx;
};
const int maxn = 2e5 + 10;
int n;
Node arr[maxn];
int b[maxn], cnt;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i].idx);
    arr[i].val = i - arr[i].idx;
  }
  std::sort(arr + 1, arr + 1 + n, [](const Node &a, const Node &b) {
    if (a.val == b.val) return a.idx < b.idx;
    return a.val < b.val;
  });
  for (int i = 1; i <= n; i++) {
    if (arr[i].val < 0) continue;
    int dex = std::lower_bound(b + 1, b + cnt + 1, arr[i].idx) - b;
    b[dex] = arr[i].idx;
    cnt = std::max(cnt, dex);
  }
  printf("%d\n", cnt);
  return 0;
}

```