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 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("");
  }
  