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;
  namespace io {
  char ibuf[(1 << 20)], *iS, *iT, obuf[(1 << 20)],
      *oS = obuf, *oT = oS + (1 << 20) - 1, c, qu[55];
  int f, qr;
  inline void flush(void) {
    return fwrite(obuf, 1, oS - obuf, stdout), oS = obuf, void();
  }
  inline char getch(void) {
    return (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, (1 << 20), stdin),
                        (iS == iT ? EOF : *iS++))
                     : *iS++);
  }
  inline void putch(char x) {
    *oS++ = x;
    if (oS == oT) flush();
    return;
  }
  string getstr(void) {
    string s = "";
    char c = getch();
    while (c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == EOF)
      c = getch();
    while (!(c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == EOF))
      s.push_back(c), c = getch();
    return s;
  }
  void putstr(string str, int begin = 0, int end = -1) {
    if (end == -1) end = str.size();
    for (register int i = begin; i < end; i++) putch(str[i]);
    return;
  }
  template <typename T>
  inline T read() {
    register T x = 0;
    for (f = 1, c = getch(); c < '0' || c > '9'; c = getch())
      if (c == '-') f = -1;
    for (x = 0; c <= '9' && c >= '0'; c = getch()) x = x * 10 + (c & 15);
    return x * f;
  }
  template <typename T>
  inline void write(const T& t) {
    register T x = t;
    if (!x) putch('0');
    if (x < 0) putch('-'), x = -x;
    while (x) qu[++qr] = x % 10 + '0', x /= 10;
    while (qr) putch(qu[qr--]);
    return;
  }
  struct Flusher_ {
    ~Flusher_() { flush(); }
  } io_flusher_;
  }  // namespace io
  using io::getch;
  using io::getstr;
  using io::putch;
  using io::putstr;
  using io::read;
  using io::write;
  bool mem1;
  pair<int, int> a[200005];
  int S[200005];
  void solve(void) {
    int n = read<int>();
    for (int i = 1; i <= n; i++) {
      int x = read<int>();
      a[i] = {i - x, x};
    }
    sort(a + 1, a + n + 1);
    int answer = 0;
    memset(S, 0x3f, sizeof(S));
    for (int i = 1; i <= n; i++) {
      if (a[i].first < 0) continue;
      int x = a[i].second, p = lower_bound(S + 1, S + answer + 1, x) - S;
      S[p] = min(S[p], x);
      answer = max(answer, p);
    }
    write(answer), putch('\n');
    return;
  }
  bool mem2;
  int main() {
    clock_t t1 = clock();
    cerr << "Memory: " << abs(&mem1 - &mem2) / 1024.0 / 1024.0 << "MB" << endl;
    int _ = 1;
    while (_--) solve();
    clock_t t2 = clock();
    cerr << "Time: " << (t2 - t1) * 1000.0 / CLOCKS_PER_SEC << "ms" << endl;
    return 0;
  }
  