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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(ios::badbit | ios::failbit);
  int n;
  cin >> n;
  vector<int> a;
  for (auto i = 0; i < n; ++i) {
    int x;
    cin >> x, --x;
    if (x <= i) {
      x -= i;
      x = -x;
      auto it = lower_bound(a.begin(), a.end(), x);
      if (it == a.end()) {
        a.push_back(x);
      } else {
        *it = x;
      }
    }
  }
  cout << (int)a.size() << "\n";
  return 0;
}
