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;
  constexpr long long mod = 998244353;
  const long long INF = mod * mod;
  const long double eps = 1e-8;
  const long double pi = acosl(-1.0);
  long long mod_pow(long long x, long long n, long long m = mod) {
    if (n < 0) {
      long long res = mod_pow(x, -n, m);
      return mod_pow(res, m - 2, m);
    }
    if (abs(x) >= m) x %= m;
    if (x < 0) x += m;
    long long res = 1;
    while (n) {
      if (n & 1) res = res * x % m;
      x = x * x % m;
      n >>= 1;
    }
    return res;
  }
  struct modint {
    long long n;
    modint() : n(0) { ; }
    modint(long long m) : n(m) {
      if (n >= mod)
        n %= mod;
      else if (n < 0)
        n = (n % mod + mod) % mod;
    }
    operator int() { return n; }
  };
  bool operator==(modint a, modint b) { return a.n == b.n; }
  modint operator+=(modint& a, modint b) {
    a.n += b.n;
    if (a.n >= mod) a.n -= mod;
    return a;
  }
  modint operator-=(modint& a, modint b) {
    a.n -= b.n;
    if (a.n < 0) a.n += mod;
    return a;
  }
  modint operator*=(modint& a, modint b) {
    a.n = ((long long)a.n * b.n) % mod;
    return a;
  }
  modint operator+(modint a, modint b) { return a += b; }
  modint operator-(modint a, modint b) { return a -= b; }
  modint operator*(modint a, modint b) { return a *= b; }
  modint operator^(modint a, long long n) {
    if (n == 0) return modint(1);
    modint res = (a * a) ^ (n / 2);
    if (n % 2) res = res * a;
    return res;
  }
  long long inv(long long a, long long p) {
    return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
  }
  modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
  modint operator/=(modint& a, modint b) {
    a = a / b;
    return a;
  }
  const int max_n = 1 << 2;
  modint fact[max_n], factinv[max_n];
  void init_f() {
    fact[0] = modint(1);
    for (int i = 0; i < max_n - 1; i++) {
      fact[i + 1] = fact[i] * modint(i + 1);
    }
    factinv[max_n - 1] = modint(1) / fact[max_n - 1];
    for (int i = max_n - 2; i >= 0; i--) {
      factinv[i] = factinv[i + 1] * modint(i + 1);
    }
  }
  modint comb(int a, int b) {
    if (a < 0 || b < 0 || a < b) return 0;
    return fact[a] * factinv[b] * factinv[a - b];
  }
  modint combP(int a, int b) {
    if (a < 0 || b < 0 || a < b) return 0;
    return fact[a] * factinv[a - b];
  }
  struct Seg2T {
   private:
    vector<int> node;
    int sz;
    vector<int> vy;
    const int init_c = -mod;
  
   public:
    int f(int a, int b) { return max(a, b); }
    void add(int y) { vy.push_back(y); }
    void complete() {
      sort(vy.begin(), vy.end());
      vy.erase(unique(vy.begin(), vy.end()), vy.end());
      int n = vy.size();
      sz = 1;
      while (sz < n) sz *= 2;
      node.resize(2 * sz - 1, init_c);
    }
    void update(int k, int c) {
      k = lower_bound(vy.begin(), vy.end(), k) - vy.begin();
      k += sz - 1;
      node[k] = f(node[k], c);
      while (k > 0) {
        k = (k - 1) / 2;
        node[k] = f(node[k * 2 + 1], node[k * 2 + 2]);
      }
    }
    int query(int a, int b, int k, int l, int r) {
      if (r <= a || b <= l) return init_c;
      if (a <= l && r <= b)
        return node[k];
      else {
        long long vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
        long long vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
        return f(vl, vr);
      }
    }
    int query(int a, int b) {
      a = lower_bound(vy.begin(), vy.end(), a) - vy.begin();
      b = lower_bound(vy.begin(), vy.end(), b) - vy.begin();
      return query(a, b, 0, 0, sz);
    }
  };
  struct SegT {
   private:
    int sz;
    vector<Seg2T> node;
    const int init_c = -mod;
  
   public:
    int f(int a, int b) { return max(a, b); }
    SegT(vector<pair<int, int> > v, int n) {
      sz = 1;
      while (sz < n) sz *= 2;
      node.resize(sz * 2 - 1);
      for (int i = 0; i < v.size(); i++) {
        int k = v[i].first, y = v[i].second;
        k += sz - 1;
        node[k].add(y);
        while (k > 0) {
          k = (k - 1) / 2;
          node[k].add(y);
        }
      }
      for (int i = 0; i < 2 * sz - 1; i++) node[i].complete();
    }
    void update(int x, int y, int c) {
      x += sz - 1;
      node[x].update(y, c);
      while (x) {
        x = (x - 1) / 2;
        node[x].update(y, c);
      }
    }
    int query(int a, int b, int x, int y, int k = 0, int l = 0, int r = -1) {
      if (r < 0) r = sz;
      if (r <= a || b <= l)
        return init_c;
      else if (a <= l && r <= b)
        return node[k].query(x, y);
      else {
        int vl = query(a, b, x, y, k * 2 + 1, l, (l + r) / 2);
        int vr = query(a, b, x, y, k * 2 + 2, (l + r) / 2, r);
        return f(vl, vr);
      }
    }
  };
  int a[1000][1000];
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<int> d(n);
    for (int i = 0; i < n; i++) d[i] = a[i] - i;
    vector<int> vd = d;
    sort((vd).begin(), (vd).end());
    vd.erase(unique((vd).begin(), (vd).end()), vd.end());
    for (int i = 0; i < n; i++)
      d[i] = lower_bound((vd).begin(), (vd).end(), d[i]) - vd.begin();
    vector<pair<int, int> > ori(n);
    for (int i = 0; i < n; i++) ori[i] = {i, d[i]};
    SegT st(ori, n);
    vector<pair<int, int> > vp;
    for (int i = 0; i < n; i++) vp.push_back({a[i], -i});
    sort((vp).begin(), (vp).end());
    for (pair<int, int> p : vp) {
      int id = -p.second;
      int val = st.query(0, id, d[id], vd.size());
      if (id + 1 >= a[id]) val = max(val, 0);
      val += 1;
      st.update(id, d[id], val);
    }
    int ans = 0;
    ans = max(ans, st.query(0, n, 0, vd.size()));
    cout << ans << "\n";
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    solve();
    return 0;
  }
  