#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long sum = 0, f = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar()) f |= (ch == '-');
  for (; isdigit(ch); ch = getchar())
    sum = ((sum << 1) + (sum << 3) + (ch ^ 48));
  return f ? -sum : sum;
}
void write(const long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
char getc() {
  char ch = getchar();
  for (; !isalpha(ch); ch = getchar())
    ;
  return ch;
}
long long n, a[550000], tot, kkk, f[550000];
struct nd {
  long long x, y;
} q[550000];
bool cmp(nd a, nd b) {
  if (a.y != b.y) return a.y < b.y;
  return a.x < b.x;
}
int main() {
  n = read();
  for (register long long i = (1); i <= (n); i++) a[i] = read();
  for (register long long i = (1); i <= (n); i++)
    if (a[i] <= i) q[++tot] = (nd){a[i], i - a[i]};
  sort(q + 1, q + tot + 1, cmp);
  for (register long long i = (1); i <= (tot); i++) a[i] = q[i].x;
  f[1] = a[1];
  kkk = 1;
  for (register long long i = (2); i <= (tot); i++) {
    if (a[i] > f[kkk])
      f[++kkk] = a[i];
    else
      f[lower_bound(f + 1, f + kkk + 1, a[i]) - f] = a[i];
  }
  write(tot == 0 ? 0 : kkk), puts("");
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long sum = 0, f = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar()) f |= (ch == '-');
  for (; isdigit(ch); ch = getchar())
    sum = ((sum << 1) + (sum << 3) + (ch ^ 48));
  return f ? -sum : sum;
}
void write(const long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
char getc() {
  char ch = getchar();
  for (; !isalpha(ch); ch = getchar())
    ;
  return ch;
}
long long n, a[550000], tot, kkk, f[550000];
struct nd {
  long long x, y;
} q[550000];
bool cmp(nd a, nd b) {
  if (a.y != b.y) return a.y < b.y;
  return a.x < b.x;
}
int main() {
  n = read();
  for (register long long i = (1); i <= (n); i++) a[i] = read();
  for (register long long i = (1); i <= (n); i++)
    if (a[i] <= i) q[++tot] = (nd){a[i], i - a[i]};
  sort(q + 1, q + tot + 1, cmp);
  for (register long long i = (1); i <= (tot); i++) a[i] = q[i].x;
  f[1] = a[1];
  kkk = 1;
  for (register long long i = (2); i <= (tot); i++) {
    if (a[i] > f[kkk])
      f[++kkk] = a[i];
    else
      f[lower_bound(f + 1, f + kkk + 1, a[i]) - f] = a[i];
  }
  write(tot == 0 ? 0 : kkk), puts("");
}

```