Problem:
Score : 100 points

Problem Statement

This contest is CODEFESTIVAL, which can be shortened to the string CF by deleting some characters.

Mr. Takahashi, full of curiosity, wondered if he could obtain CF from other strings in the same way.

You are given a string s consisting of uppercase English letters.
Determine whether the string CF can be obtained from the string s by deleting some characters.

Constraints

2 ≤ |s| ≤ 100

All characters in s are uppercase English letters (A-Z).

Input

The input is given from Standard Input in the following format:

s

Output

Print Yes if the string CF can be obtained from the string s by deleting some characters.
Otherwise print No.

Sample Input 1

CODEFESTIVAL

Sample Output 1

Yes

CF is obtained by deleting characters other than the first character C and the fifth character F.

Sample Input 2

FESTIVALCODE

Sample Output 2

No

FC can be obtained but CF cannot be obtained because you cannot change the order of the characters.

Sample Input 3

CF

Sample Output 3

Yes

It is also possible not to delete any characters.

Sample Input 4

FCF

Sample Output 4

Yes

CF is obtained by deleting the first character.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.util.function.BiFunction;
import java.util.function.Function;

public class Main {
  void run() {
    String str = sc.next();
    System.out.println(str.indexOf('C') >= 0 && str.indexOf('F') >= 0 ? "Yes" : "No");
  }

  Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {
    new Main().run();
  }

  int ni() {
    return Integer.parseInt(sc.next());
  }

  void debug(Object... os) {
    System.err.println(Arrays.deepToString(os));
  }

  class BIT<T> {
    int n;
    ArrayList<T> bit;
    BiFunction<T, T, T> bif;

    BIT(int n, BiFunction<T, T, T> bif, T defaultValue) {
      this.n = n;
      bit = new ArrayList<>(n + 1);
      for (int i = 0; i < n + 1; ++i) {
        bit.add(defaultValue);
      }
      this.bif = bif;
    }

    void update(int i, T v) {
      for (int x = i; x <= n; x += x & -x) {
        bit.set(x, bif.apply(bit.get(x), v));
      }
    }

    T reduce(int i, T defaultValue) {
      T ret = defaultValue;
      for (int x = i; x > 0; x -= x & -x) {
        ret = bif.apply(ret, bit.get(x));
      }
      return ret;
    }
  }

  long MOD = 1_000_000_007;

  long pow(long a, long r) {
    long sum = 1;
    while (r > 0) {
      if ((r & 1) == 1) {
        sum *= a;
        sum %= MOD;
      }
      a *= a;
      a %= MOD;
      r >>= 1;
    }
    return sum;
  }

  long C(int n, int r) {
    long sum = 1;
    for (int i = n; 0 < i; --i) {
      sum *= i;
      sum %= MOD;
    }
    long s = 1;
    for (int i = r; 0 < i; --i) {
      s *= i;
      s %= MOD;
    }
    sum *= pow(s, MOD - 2);
    sum %= MOD;

    long t = 1;
    for (int i = n - r; 0 < i; --i) {
      t *= i;
      t %= MOD;
    }
    sum *= pow(t, MOD - 2);
    sum %= MOD;

    return sum;
  }

  double GOLDEN_RATIO = (1.0 + Math.sqrt(5)) / 2.0;

  /**
   * 黄金分割探索
   *
   * @param left  下限
   * @param right 上限
   * @param f     探索する関数
   * @param comp  上に凸な関数を探索するときは、Comparator.comparingDouble(Double::doubleValue)
   *              下に凸な関数を探索するときは、Comparator.comparingDouble(Double::doubleValue).reversed()
   * @return 極値の座標x
   */
  double goldenSectionSearch(double left, double right, Function<Double, Double> f, Comparator<Double> comp) {
    double c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
    double c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
    double d1 = f.apply(c1);
    double d2 = f.apply(c2);
    while (right - left > 1e-9) {
      if (comp.compare(d1, d2) > 0) {
        right = c2;
        c2 = c1;
        d2 = d1;
        c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
        d1 = f.apply(c1);
      } else {
        left = c1;
        c1 = c2;
        d1 = d2;
        c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
        d2 = f.apply(c2);
      }
    }
    return right;
  }

  /**
   * [a,b]をm:nに内分する点を返す
   */
  double divideInternally(double a, double b, double m, double n) {
    return (n * a + m * b) / (m + n);
  }
}