#include <bits/stdc++.h>
using namespace std;
int tree[4 * 200005];
void update(int n, int st, int en, int index, int val) {
  if (st == en) {
    tree[n] = val;
    return;
  }
  int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
  if (index <= mid)
    update(lc, st, mid, index, val);
  else
    update(rc, mid + 1, en, index, val);
  tree[n] = max(tree[lc], tree[rc]);
}
int query(int n, int st, int en, int l, int r) {
  if (r < st || en < l) return 0;
  if (l <= st && en <= r) return tree[n];
  int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
  return max(query(lc, st, mid, l, r), query(rc, mid + 1, en, l, r));
}
vector<int> pos[200005 + 5];
int main() {
  int n, inp;
  scanf("%d", &n);
  int mx = 0;
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &inp);
    pos[inp].push_back(i);
  }
  for (int num = 1; num < 200005; num++) {
    reverse(pos[num].begin(), pos[num].end());
    for (auto position : pos[num]) {
      if (position < num) continue;
      int highestDeletePossible = position - num;
      int maxAns = query(1, 0, 200005, 0, highestDeletePossible) + 1;
      update(1, 0, 200005, highestDeletePossible, maxAns);
    }
  }
  cout << query(1, 0, 200005, 0, 200005) << '\n';
  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>
using namespace std;
int tree[4 * 200005];
void update(int n, int st, int en, int index, int val) {
  if (st == en) {
    tree[n] = val;
    return;
  }
  int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
  if (index <= mid)
    update(lc, st, mid, index, val);
  else
    update(rc, mid + 1, en, index, val);
  tree[n] = max(tree[lc], tree[rc]);
}
int query(int n, int st, int en, int l, int r) {
  if (r < st || en < l) return 0;
  if (l <= st && en <= r) return tree[n];
  int mid = (st + en) / 2, lc = 2 * n, rc = lc + 1;
  return max(query(lc, st, mid, l, r), query(rc, mid + 1, en, l, r));
}
vector<int> pos[200005 + 5];
int main() {
  int n, inp;
  scanf("%d", &n);
  int mx = 0;
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &inp);
    pos[inp].push_back(i);
  }
  for (int num = 1; num < 200005; num++) {
    reverse(pos[num].begin(), pos[num].end());
    for (auto position : pos[num]) {
      if (position < num) continue;
      int highestDeletePossible = position - num;
      int maxAns = query(1, 0, 200005, 0, highestDeletePossible) + 1;
      update(1, 0, 200005, highestDeletePossible, maxAns);
    }
  }
  cout << query(1, 0, 200005, 0, 200005) << '\n';
  return 0;
}

```