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 read() {
    int x = 0, f = 0;
    char c = getchar();
    while (c < '0') f |= (c == '-'), c = getchar();
    while (c >= '0') x = x * 10 + c - '0', c = getchar();
    return f ? -x : x;
  }
  int Ans, n, a[500005], ans[500005], cnt, b[500005];
  void solve(int l, int r) {
    if (l >= r) return;
    int mid = (l + r) / 2;
    solve(l, mid);
    sort(b + mid + 1, b + r + 1, [](int x, int y) { return a[x] < a[y]; });
    int ma = 0, j = mid + 1;
    for (int i = l; i <= mid; i++) {
      while (j <= r && a[b[j]] <= a[b[i]]) ans[b[j]] = max(ans[b[j++]], ma + 1);
      ma = max(ans[b[i]], ma);
    }
    while (j <= r) ans[b[j]] = max(ans[b[j++]], ma + 1);
    sort(b + mid + 1, b + r + 1, [](int x, int y) {
      if (x - a[x] == y - a[y]) return a[x] < a[y];
      return x - a[x] < y - a[y];
    });
    solve(mid + 1, r);
    sort(b + l, b + r + 1, [](int x, int y) { return a[x] < a[y]; });
  }
  int main() {
    n = read();
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      ans[i] = 1;
      if (a[i] <= i) b[++cnt] = i;
    }
    sort(b + 1, b + cnt + 1, [](int x, int y) {
      if (x - a[x] == y - a[y]) return a[x] < a[y];
      return x - a[x] < y - a[y];
    });
    solve(1, cnt);
    for (int i = 1; i <= n; i++) Ans = max(Ans, ans[i]);
    cout << min(Ans, cnt);
    return 0;
  }
  