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>
#pragma GCC optimize("-O3")
using namespace std;
const long long mod = 998244353;
const int N = 2e5 + 5;
int n, a[N], bit[N];
void ins(int i, int x) {
  while (i <= n) {
    bit[i] = max(bit[i], x);
    i += (i & -i);
  }
}
int mx(int i) {
  int ans = 0;
  while (i > 0) {
    ans = max(ans, bit[i]);
    i = (i & i - 1);
  }
  return ans;
}
void solve() {
  cin >> n;
  for (int i = (int)0; i != (int)n; i = i + 1) cin >> a[i];
  ;
  int ans = 0;
  fill(bit, bit + n + 1, 0);
  vector<int> v[n + 1];
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] <= n) v[a[i]].push_back(i + 1);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = (int)v[i].size() - 1; j >= 0; j--) {
      if (v[i][j] - i < 0) continue;
      int dp = 1;
      dp = max(dp, 1 + mx(v[i][j] + 1 - i));
      ans = max(ans, dp);
      ins(v[i][j] + 1 - i, dp);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}
