#include <bits/stdc++.h>
using namespace std;
int n;
int a[200007];
int c[1000007];
vector<pair<int, int> > p;
void add(int v, int val) {
  v += 200000;
  for (int i = v; i <= 400007; i += ((i) & (-i))) {
    c[i] = max(c[i], val);
  }
  return;
}
int ask(int v) {
  v += 200000;
  int res = 0;
  for (int i = v; i; i -= ((i) & (-i))) {
    res = max(res, c[i]);
  }
  return res;
}
vector<int> te[200007];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] > n || a[i] > i) continue;
    p.emplace_back(a[i], i - a[i]);
  }
  int ans = 0;
  sort(p.begin(), p.end());
  for (auto [x, y] : p) {
    te[x].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    vector<pair<int, int> > q;
    for (auto y : te[i]) {
      int tp = ask(y) + 1;
      ans = max(ans, tp);
      q.emplace_back(y, tp);
    }
    for (auto [y, tp] : q) {
      add(y, tp);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int n;
int a[200007];
int c[1000007];
vector<pair<int, int> > p;
void add(int v, int val) {
  v += 200000;
  for (int i = v; i <= 400007; i += ((i) & (-i))) {
    c[i] = max(c[i], val);
  }
  return;
}
int ask(int v) {
  v += 200000;
  int res = 0;
  for (int i = v; i; i -= ((i) & (-i))) {
    res = max(res, c[i]);
  }
  return res;
}
vector<int> te[200007];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] > n || a[i] > i) continue;
    p.emplace_back(a[i], i - a[i]);
  }
  int ans = 0;
  sort(p.begin(), p.end());
  for (auto [x, y] : p) {
    te[x].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    vector<pair<int, int> > q;
    for (auto y : te[i]) {
      int tp = ask(y) + 1;
      ans = max(ans, tp);
      q.emplace_back(y, tp);
    }
    for (auto [y, tp] : q) {
      add(y, tp);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```