#include <bits/stdc++.h>
template <class T>
void read(T &a) {
  a = 0;
  char x = getchar();
  bool f = 0;
  for (; x < '0' || x > '9'; x = getchar()) f |= x == '-';
  for (; x >= '0' && x <= '9'; x = getchar()) a = (a << 3) + (a << 1) + x - '0';
  if (f) a = -a;
}
template <class T, class... Args>
void read(T &a, Args &...args) {
  read(a);
  read(args...);
}
using namespace std;
const int N = 2e5 + 5;
int n;
int f[N], lim;
pair<int, int> d[N];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first ^ b.first) {
    return a.first < b.first;
  }
  return a.second > b.second;
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(d[i].first);
    d[i].second = i;
  }
  sort(d + 1, d + n + 1, cmp);
  int lim = 0;
  for (int i = 1; i <= n; ++i) {
    int c = d[i].second - d[i].first;
    if (c < 0) continue;
    if (lim == 0) {
      f[++lim] = c;
      continue;
    }
    if (c >= f[lim]) {
      f[++lim] = c;
      continue;
    }
    int l = 1, r = lim + 1;
    while (l < r - 1) {
      int mid = l + r >> 1;
      if (f[mid] <= c) {
        l = mid;
      } else {
        r = mid;
      }
    }
    if (f[l] > c) {
      f[l] = c;
    } else if (l != lim) {
      f[l + 1] = c;
    }
  }
  printf("%d\n", lim);
  return 0;
}

### Prompt

Your challenge is to write 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>
template <class T>
void read(T &a) {
  a = 0;
  char x = getchar();
  bool f = 0;
  for (; x < '0' || x > '9'; x = getchar()) f |= x == '-';
  for (; x >= '0' && x <= '9'; x = getchar()) a = (a << 3) + (a << 1) + x - '0';
  if (f) a = -a;
}
template <class T, class... Args>
void read(T &a, Args &...args) {
  read(a);
  read(args...);
}
using namespace std;
const int N = 2e5 + 5;
int n;
int f[N], lim;
pair<int, int> d[N];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.first ^ b.first) {
    return a.first < b.first;
  }
  return a.second > b.second;
}
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(d[i].first);
    d[i].second = i;
  }
  sort(d + 1, d + n + 1, cmp);
  int lim = 0;
  for (int i = 1; i <= n; ++i) {
    int c = d[i].second - d[i].first;
    if (c < 0) continue;
    if (lim == 0) {
      f[++lim] = c;
      continue;
    }
    if (c >= f[lim]) {
      f[++lim] = c;
      continue;
    }
    int l = 1, r = lim + 1;
    while (l < r - 1) {
      int mid = l + r >> 1;
      if (f[mid] <= c) {
        l = mid;
      } else {
        r = mid;
      }
    }
    if (f[l] > c) {
      f[l] = c;
    } else if (l != lim) {
      f[l + 1] = c;
    }
  }
  printf("%d\n", lim);
  return 0;
}

```