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 maxn = 2 * 1e5 + 10, maxm = 2 * 8 * maxn, mod = 998244353;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1), eps = 1e-7;
int n;
int a[maxn];
int p[maxn];
int f[maxn];
int tr[maxn];
int lowbit(int x) { return x & (-x); }
void modify(int x, int k) {
  while (x <= n) {
    tr[x] = max(tr[x], k);
    x += lowbit(x);
  }
}
int query(int x) {
  int res = 0;
  while (x) {
    res = max(res, tr[x]);
    x -= lowbit(x);
  }
  return res;
}
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];
  }
  for (int i = 1; i <= n; i++) {
    p[i] = i - a[i];
    if (p[i] < 0) p[i] = -1;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (p[i] == -1) continue;
    int t = query(p[i] + 1);
    f[i] = t + 1;
    modify(p[i] + 1, f[i]);
    ans = max(ans, f[i]);
  }
  cout << ans << "\n";
  return 0;
}
