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 inf = 1 << 30;
const long long Inf = 1ll << 60ll;
namespace io {
char buf[(1 << 23)], *p1 = buf, *p2 = buf, c;
int f;
template <typename T>
T read() {
  T x = 0;
  f = 0;
  c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (!isdigit(c))
    (c == '-') && (f = 1),
        c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin),
                          p1 == p2)
                 ? EOF
                 : *p1++);
  while (isdigit(c))
    x = x * 10 + (c & 15),
    c = (p1 == p2 &&
                 (p2 = (p1 = buf) + fread(buf, 1, (1 << 23), stdin), p1 == p2)
             ? EOF
             : *p1++);
  return f ? -x : x;
}
}  // namespace io
const int N = 200005;
struct node {
  int a, b;
} p[N];
int n, c[N], t[N];
bool cmp(node A, node B) { return A.a < B.a; }
void update(int x, int v) {
  while (x <= n) {
    t[x] = max(t[x], v);
    x += x & (-x);
  }
}
int query(int x) {
  int res = 0;
  while (x) {
    res = max(res, t[x]);
    x -= x & (-x);
  }
  return res;
}
int main() {
  n = io::read<int>();
  for (int i = 1; i <= n; ++i) {
    p[i].b = io::read<int>();
    p[i].a = i - p[i].b;
  }
  sort(p + 1, p + n + 1, cmp);
  int tot = 0;
  for (int i = 1; i <= n; ++i) {
    if (p[i].a >= 0) c[++tot] = p[i].b;
  }
  int ans = 0;
  for (int i = 1; i <= tot; ++i) {
    int res = query(c[i] + 1);
    ans = max(ans, res + 1);
    update(c[i] + 1, res + 1);
  }
  printf("%d\n", ans);
  return 0;
}
