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;
const int N = 2e5 + 5;
int n, f[N], tot = 0;
struct node {
  int x, y;
} a[N];
bool cmp(node xx, node yy) {
  if (xx.x != yy.x)
    return xx.x < yy.x;
  else
    return xx.y < yy.y;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].x;
    a[i].y = i - a[i].x;
  }
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; ++i) {
    if (a[i].y < 0) continue;
    if (a[i].y >= f[tot])
      f[++tot] = a[i].y;
    else {
      int p = lower_bound(f + 1, f + tot + 1, a[i].y) - f;
      f[p] = a[i].y;
    }
  }
  cout << tot << endl;
  return 0;
}
