#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 0;
  char c = getchar();
  while (c < '0') f |= (c == '-'), c = getchar();
  while (c >= '0') x = x * 10 + c - '0', c = getchar();
  return f ? -x : x;
}
int Ans, n, a[500005], ans[500005], cnt, b[500005];
void solve(int l, int r) {
  if (l >= r) return;
  int mid = (l + r) / 2;
  solve(l, mid);
  sort(b + mid + 1, b + r + 1, [](int x, int y) { return a[x] < a[y]; });
  int ma = 0, j = mid + 1;
  for (int i = l; i <= mid; i++) {
    while (j <= r && a[b[j]] <= a[b[i]]) ans[b[j]] = max(ans[b[j++]], ma + 1);
    ma = max(ans[b[i]], ma);
  }
  while (j <= r) ans[b[j]] = max(ans[b[j++]], ma + 1);
  sort(b + mid + 1, b + r + 1, [](int x, int y) {
    if (x - a[x] == y - a[y]) return a[x] < a[y];
    return x - a[x] < y - a[y];
  });
  solve(mid + 1, r);
  sort(b + l, b + r + 1, [](int x, int y) { return a[x] < a[y]; });
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    ans[i] = 1;
    if (a[i] <= i) b[++cnt] = i;
  }
  sort(b + 1, b + cnt + 1, [](int x, int y) {
    if (x - a[x] == y - a[y]) return a[x] < a[y];
    return x - a[x] < y - a[y];
  });
  solve(1, cnt);
  for (int i = 1; i <= n; i++) Ans = max(Ans, ans[i]);
  cout << min(Ans, cnt);
  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;
inline int read() {
  int x = 0, f = 0;
  char c = getchar();
  while (c < '0') f |= (c == '-'), c = getchar();
  while (c >= '0') x = x * 10 + c - '0', c = getchar();
  return f ? -x : x;
}
int Ans, n, a[500005], ans[500005], cnt, b[500005];
void solve(int l, int r) {
  if (l >= r) return;
  int mid = (l + r) / 2;
  solve(l, mid);
  sort(b + mid + 1, b + r + 1, [](int x, int y) { return a[x] < a[y]; });
  int ma = 0, j = mid + 1;
  for (int i = l; i <= mid; i++) {
    while (j <= r && a[b[j]] <= a[b[i]]) ans[b[j]] = max(ans[b[j++]], ma + 1);
    ma = max(ans[b[i]], ma);
  }
  while (j <= r) ans[b[j]] = max(ans[b[j++]], ma + 1);
  sort(b + mid + 1, b + r + 1, [](int x, int y) {
    if (x - a[x] == y - a[y]) return a[x] < a[y];
    return x - a[x] < y - a[y];
  });
  solve(mid + 1, r);
  sort(b + l, b + r + 1, [](int x, int y) { return a[x] < a[y]; });
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    ans[i] = 1;
    if (a[i] <= i) b[++cnt] = i;
  }
  sort(b + 1, b + cnt + 1, [](int x, int y) {
    if (x - a[x] == y - a[y]) return a[x] < a[y];
    return x - a[x] < y - a[y];
  });
  solve(1, cnt);
  for (int i = 1; i <= n; i++) Ans = max(Ans, ans[i]);
  cout << min(Ans, cnt);
  return 0;
}

```