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;
  const long long inf = 1e18;
  long long n, lis[200069];
  pair<long long, long long> a[200069];
  int main() {
    long long i, k, p, z = 0;
    scanf("%lld", &n);
    for (i = 1; i <= n; i++) {
      scanf("%lld", &k);
      k = i - k;
      a[i] = {i - k, -k};
      lis[i] = inf;
    }
    sort(a + 1, a + n + 1);
    for (i = 1; i <= n; i++) {
      k = -a[i].second;
      if (a[i].first > 0 && k >= 0) {
        p = upper_bound(lis + 1, lis + n + 1, k) - lis;
        z = max(z, p);
        lis[p] = k;
      }
    }
    printf("%lld\n", z);
  }
  