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 q[10000000];
int p[10000000];
struct node {
  int num;
  int c;
} a[10000000];
bool cmp(node x1, node x2) {
  if (x1.num < x2.num) {
    return 1;
  } else if (x1.num == x2.num) {
    return x1.c < x2.c;
  } else {
    return 0;
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].num;
    a[i].c = a[i].num - i;
  }
  int ans = 1;
  sort(a + 1, a + n + 1, cmp);
  q[1] = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i].c <= q[ans]) {
      ans++;
      q[ans] = a[i].c;
      p[ans] = a[i].num;
    } else {
      int tem = lower_bound(q + 2, q + ans + 1, a[i].c, greater<int>()) - q;
      if (tem != 1 && tem != ans + 1) {
        q[tem] = a[i].c;
        p[tem] = a[i].num;
      }
    }
  }
  cout << ans - 1;
  return 0;
}
