#include <bits/stdc++.h>
using namespace std;
bool isprime(long long k) {
  for (long long i = 2; i <= sqrt(k); i++)
    if (k % i == 0) return false;
  return true;
}
long long bm(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  long long t = bm(a, b / 2, mod);
  t = t * t % mod;
  return (b % 2 == 1 ? t * a % mod : t);
}
long long inv(long long a, long long mod) { return bm(a, mod - 2, mod); }
void gay(long long TC) { cout << "Case #" << TC << ": "; }
struct pe {
  long long v;
  long long id;
} a[200005];
long long b[200005];
bool cmp(pe x, pe y) {
  if (x.v == y.v) return x.id < y.id;
  return x.v < y.v;
}
long long d[200005];
long long cnt;
long long n;
bool can(long long k) {
  long long pos;
  cnt = 0;
  for (long long i = 0; i < n; i++) d[i] = 0;
  for (long long i = 1; i < n; i++) {
    if (a[i].v >= 0) {
      pos = i;
      break;
    }
  }
  for (long long i = pos; i < n; i++) {
    if (cnt == 0 || a[i].id > d[cnt]) {
      d[++cnt] = a[i].id;
    } else {
      long long k = lower_bound(d + 1, d + 1 + cnt, a[i].id) - d;
      if (k >= 0) d[k] = a[i].id;
    }
    if (cnt >= k) return true;
  }
  return false;
}
void solve(long long TC) {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> b[i];
    if (b[i] > n) {
      b[i] = n + 1;
    }
    a[i].v = i - b[i];
    a[i].id = b[i];
  }
  sort(a + 1, a + 1 + n, cmp);
  long long l = 0, r = n;
  n++;
  while (l < r) {
    long long m = (l + r + 1) / 2;
    if (!can(m))
      r = m - 1;
    else
      l = m;
  }
  cout << l << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 1; i <= t; i++) solve(i);
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
bool isprime(long long k) {
  for (long long i = 2; i <= sqrt(k); i++)
    if (k % i == 0) return false;
  return true;
}
long long bm(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  long long t = bm(a, b / 2, mod);
  t = t * t % mod;
  return (b % 2 == 1 ? t * a % mod : t);
}
long long inv(long long a, long long mod) { return bm(a, mod - 2, mod); }
void gay(long long TC) { cout << "Case #" << TC << ": "; }
struct pe {
  long long v;
  long long id;
} a[200005];
long long b[200005];
bool cmp(pe x, pe y) {
  if (x.v == y.v) return x.id < y.id;
  return x.v < y.v;
}
long long d[200005];
long long cnt;
long long n;
bool can(long long k) {
  long long pos;
  cnt = 0;
  for (long long i = 0; i < n; i++) d[i] = 0;
  for (long long i = 1; i < n; i++) {
    if (a[i].v >= 0) {
      pos = i;
      break;
    }
  }
  for (long long i = pos; i < n; i++) {
    if (cnt == 0 || a[i].id > d[cnt]) {
      d[++cnt] = a[i].id;
    } else {
      long long k = lower_bound(d + 1, d + 1 + cnt, a[i].id) - d;
      if (k >= 0) d[k] = a[i].id;
    }
    if (cnt >= k) return true;
  }
  return false;
}
void solve(long long TC) {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> b[i];
    if (b[i] > n) {
      b[i] = n + 1;
    }
    a[i].v = i - b[i];
    a[i].id = b[i];
  }
  sort(a + 1, a + 1 + n, cmp);
  long long l = 0, r = n;
  n++;
  while (l < r) {
    long long m = (l + r + 1) / 2;
    if (!can(m))
      r = m - 1;
    else
      l = m;
  }
  cout << l << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 1; i <= t; i++) solve(i);
  return 0;
}

```