#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 200005;
int n;
pair<int, int> a[N];
int maxu[N * 10];
void ubd(int tl, int tr, int x, int y, int pos) {
  if (tl == tr) {
    maxu[pos] = max(maxu[pos], y);
    return;
  }
  int m = (tl + tr) / 2;
  if (x <= m)
    ubd(tl, m, x, y, pos * 2);
  else
    ubd(m + 1, tr, x, y, pos * 2 + 1);
  maxu[pos] = max(maxu[pos * 2], maxu[pos * 2 + 1]);
}
int qry(int tl, int tr, int l, int r, int pos) {
  if (l > r) return -N;
  if (tl == l && tr == r) return maxu[pos];
  int m = (tl + tr) / 2;
  return max(qry(tl, m, l, min(m, r), pos * 2),
             qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
}
void solv() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i].first;
  for (int i = 1; i <= n; ++i) {
    a[i].second = i;
    a[i].first = (i - a[i].first);
  }
  sort(a + 1, a + n + 1);
  for (int i = 0; i < N * 10; ++i) maxu[i] = -N;
  ubd(0, N * 2, N + 0, 0, 1);
  for (int i = 1; i <= n; ++i) {
    if (a[i].first < 0) continue;
    ubd(0, N * 2, N + a[i].first - a[i].second,
        qry(0, N * 2, N + a[i].first - a[i].second + 1, N * 2, 1) + 1, 1);
  }
  cout << maxu[1] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  while (tt--) solv();
  return 0;
}

### Prompt

Your challenge is to write 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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 200005;
int n;
pair<int, int> a[N];
int maxu[N * 10];
void ubd(int tl, int tr, int x, int y, int pos) {
  if (tl == tr) {
    maxu[pos] = max(maxu[pos], y);
    return;
  }
  int m = (tl + tr) / 2;
  if (x <= m)
    ubd(tl, m, x, y, pos * 2);
  else
    ubd(m + 1, tr, x, y, pos * 2 + 1);
  maxu[pos] = max(maxu[pos * 2], maxu[pos * 2 + 1]);
}
int qry(int tl, int tr, int l, int r, int pos) {
  if (l > r) return -N;
  if (tl == l && tr == r) return maxu[pos];
  int m = (tl + tr) / 2;
  return max(qry(tl, m, l, min(m, r), pos * 2),
             qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
}
void solv() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i].first;
  for (int i = 1; i <= n; ++i) {
    a[i].second = i;
    a[i].first = (i - a[i].first);
  }
  sort(a + 1, a + n + 1);
  for (int i = 0; i < N * 10; ++i) maxu[i] = -N;
  ubd(0, N * 2, N + 0, 0, 1);
  for (int i = 1; i <= n; ++i) {
    if (a[i].first < 0) continue;
    ubd(0, N * 2, N + a[i].first - a[i].second,
        qry(0, N * 2, N + a[i].first - a[i].second + 1, N * 2, 1) + 1, 1);
  }
  cout << maxu[1] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  while (tt--) solv();
  return 0;
}

```