#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
using namespace std;
template <typename val1>
void __f(const char* name, val1&& value) {
  cout << name << " = " << value << endl;
}
template <typename val1, typename... values>
void __f(const char* names, val1&& value, values&&... multiplevalue) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " = " << value << " ";
  __f(comma + 1, multiplevalue...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> trees(2 * n, 0);
  int arr[n + 1];
  function<int(int)> query = [&](int a) {
    int ans = 0;
    while (a) {
      ans = max(trees[a], ans);
      a = a - (a & (-a));
    }
    return ans;
  };
  function<void(int, int)> update = [&](int a, int v) {
    while (a < 2 * n) {
      trees[a] = max(trees[a], v);
      a = a + (a & (-a));
    }
  };
  int ans = 0;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  vector<int> index(n);
  iota(index.begin(), index.end(), 1);
  sort(index.begin(), index.end(), [&](int a, int b) {
    if (arr[a] == arr[b]) return a > b;
    return arr[a] < arr[b];
  });
  for (int i = 0; i < n; i++) {
    if (index[i] < arr[index[i]]) continue;
    int off = index[i] - arr[index[i]] + 1;
    int fc = query(off);
    ans = max(ans, fc + 1);
    update(off, fc + 1);
  }
  cout << ans << endl;
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
using namespace std;
template <typename val1>
void __f(const char* name, val1&& value) {
  cout << name << " = " << value << endl;
}
template <typename val1, typename... values>
void __f(const char* names, val1&& value, values&&... multiplevalue) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " = " << value << " ";
  __f(comma + 1, multiplevalue...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> trees(2 * n, 0);
  int arr[n + 1];
  function<int(int)> query = [&](int a) {
    int ans = 0;
    while (a) {
      ans = max(trees[a], ans);
      a = a - (a & (-a));
    }
    return ans;
  };
  function<void(int, int)> update = [&](int a, int v) {
    while (a < 2 * n) {
      trees[a] = max(trees[a], v);
      a = a + (a & (-a));
    }
  };
  int ans = 0;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  vector<int> index(n);
  iota(index.begin(), index.end(), 1);
  sort(index.begin(), index.end(), [&](int a, int b) {
    if (arr[a] == arr[b]) return a > b;
    return arr[a] < arr[b];
  });
  for (int i = 0; i < n; i++) {
    if (index[i] < arr[index[i]]) continue;
    int off = index[i] - arr[index[i]] + 1;
    int fc = query(off);
    ans = max(ans, fc + 1);
    update(off, fc + 1);
  }
  cout << ans << endl;
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}

```