description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  inline int rd() {
    int x = 0;
    char ch, t = 0;
    while (!isdigit(ch = getchar())) t |= ch == '-';
    while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
    return x = t ? -x : x;
  }
  inline long long maxll(long long x, long long y) { return x > y ? x : y; }
  inline long long minll(long long x, long long y) { return x < y ? x : y; }
  inline long long absll(long long x) { return x > 0ll ? x : -x; }
  inline long long gcd(long long x, long long y) {
    return (y == 0) ? x : gcd(y, x % y);
  }
  int n, ans;
  int minn[200005];
  pair<int, int> a[200005];
  int main() {
    n = rd();
    for (int i = 1; i <= n; i++) a[i] = pair<int, int>(rd(), -i);
    sort(a + 1, a + n + 1);
    for (int i = 1, pos, x; i <= n; i++) {
      x = -a[i].second - a[i].first;
      if (x < 0) continue;
      pos = upper_bound(minn + 1, minn + ans + 1, x) - minn;
      minn[pos] = x, ans = max(ans, pos);
    }
    printf("%d\n", ans);
    return 0;
  }
  