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;
void print(bool n) {
  if (n)
    cout << "YES";
  else
    cout << "NO";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long a[n];
  for (auto &x : a) cin >> x;
  ;
  for (auto &x : a) --x;
  vector<long long> v;
  for (long long i = 0; i < n; i++) {
    if (i - a[i] >= 0) v.push_back(a[i] - i);
  }
  vector<long long> z;
  for (auto x : v) {
    if (z.empty() || x >= z.back()) {
      z.push_back(x);
      continue;
    }
    auto it = upper_bound((z).begin(), (z).end(), x);
    if (*it > x) {
      *it = x;
    }
  }
  cout << z.size();
  return 0;
}
