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 char nc() {
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  inline int read() {
    register int x = 0, f = 1;
    register char ch = nc();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = nc();
    }
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = nc();
    return x * f;
  }
  inline void write(register int x) {
    if (!x) putchar('0');
    if (x < 0) x = -x, putchar('-');
    static int sta[20];
    register int tot = 0;
    while (x) sta[tot++] = x % 10, x /= 10;
    while (tot) putchar(sta[--tot] + 48);
  }
  int n, a[200005], f[200005], ans;
  struct node {
    int ls, rs, mx;
  } tr[200005 * 30];
  int rt, tot;
  void modify(int &x, int l, int r, int pos, int val) {
    if (!x) x = ++tot;
    tr[x].mx = max(tr[x].mx, val);
    if (l == r) return;
    int mid = l + r >> 1;
    if (pos <= mid)
      modify(tr[x].ls, l, mid, pos, val);
    else
      modify(tr[x].rs, mid + 1, r, pos, val);
  }
  int qry(int x, int l, int r, int L, int R) {
    if (!x) return 0;
    if (L <= l && r <= R) return tr[x].mx;
    int res = 0, mid = l + r >> 1;
    if (L <= mid) res = max(res, qry(tr[x].ls, l, mid, L, R));
    if (R > mid) res = max(res, qry(tr[x].rs, mid + 1, r, L, R));
    return res;
  }
  pair<int, int> p[200005];
  int kk = 0;
  void solve(int l, int r) {
    if (l == r) {
      if (l >= a[l]) f[l] = max(f[l], 1);
      return;
    }
    int mid = l + r >> 1;
    solve(l, mid);
    kk = 0;
    for (int i = l; i <= r; ++i) p[++kk] = make_pair(a[i], -i);
    sort(p + 1, p + 1 + kk);
    for (int i = 1; i <= kk; ++i) {
      int pos = -p[i].second;
      if (pos <= mid)
        modify(rt, -2e5, 2e5, a[pos] - pos, f[pos]);
      else {
        int val = qry(rt, -2e5, 2e5, a[pos] - pos, 2e5);
        if (val == 0) {
          if (pos >= a[pos]) f[pos] = max(f[pos], 1);
        } else
          f[pos] = max(f[pos], val + 1);
      }
    }
    rt = 0;
    for (int i = 1; i <= tot; ++i) tr[i].ls = tr[i].rs = tr[i].mx = 0;
    tot = 0;
    solve(mid + 1, r);
  }
  int main() {
    n = read();
    for (int i = 1; i <= n; ++i) a[i] = read();
    solve(1, n);
    for (int i = 1; i <= n; ++i) ans = max(ans, f[i]);
    printf("%d\n", ans);
    return 0;
  }
  