Problem:
Score : 200 points

Problem Statement

There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.

Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.

A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.

An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.

A string S is assigned indicating attributes of all participants. If the i-th character of string S is a, this means the participant ranked i-th in the Qualification contests is a Japanese student; b means the participant ranked i-th is an overseas student; and c means the participant ranked i-th is neither of these.

Write a program that outputs for all the participants in descending rank either Yes if they passed the Qualification contests or No if they did not pass.

Constraints

1≦N,A,B≦100000

A+B≦N

S is N characters long.

S consists only of the letters a, b and c.

Input

Inputs are provided from Standard Input in the following form.

N A B
S

Output

Output N lines. On the i-th line, output Yes if the i-th participant passed the Qualification contests or No if that participant did not pass.

Sample Input 1

10 2 3
abccabaabb

Sample Output 1

Yes
Yes
No
No
Yes
Yes
Yes
No
No
No

The first, second, fifth, sixth, and seventh participants pass the Qualification contests.

Sample Input 2

12 5 2
cabbabaacaba

Sample Output 2

No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No

The sixth participant is third among overseas students and thus does not pass the Qualification contests.

Sample Input 3

5 2 2
ccccc

Sample Output 3

No
No
No
No
No

Solution in Rust (1.15.1) (judge verdict: Accepted)

macro_rules! input {
  (source = $s:expr, $($r:tt)*) => {
      let mut iter = $s.split_whitespace();
      let mut next = || { iter.next().unwrap() };
      input_inner!{next, $($r)*}
  };
  ($($r:tt)*) => {
      let stdin = std::io::stdin();
      let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
      let mut next = move || -> String{
          bytes
              .by_ref()
              .map(|r|r.unwrap() as char)
              .skip_while(|c|c.is_whitespace())
              .take_while(|c|!c.is_whitespace())
              .collect()
      };
      input_inner!{next, $($r)*}
  };
}

macro_rules! input_inner {
  ($next:expr) => {};
  ($next:expr, ) => {};

  ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
      let $var = read_value!($next, $t);
      input_inner!{$next $($r)*}
  };

  ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
    let mut $var = read_value!($next, $t);
    input_inner!{$next $($r)*}
  };
}

macro_rules! read_value {
  ($next:expr, ( $($t:tt),* )) => {
      ( $(read_value!($next, $t)),* )
  };

  ($next:expr, [ $t:tt ; $len:expr ]) => {
      (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
  };

  ($next:expr, chars) => {
      read_value!($next, String).chars().collect::<Vec<char>>()
  };

  ($next:expr, usize1) => {
      read_value!($next, usize) - 1
  };

  ($next:expr, $t:ty) => {
      $next().parse::<$t>().expect("Parse error")
  };
}

pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T]
where
    T: Ord,
{
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i - 1] {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i - 1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }

    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] <= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();

        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j - 1] < self[i - 1] {
            j -= 1;
        }

        // Step 4: Swap that element with the pivot
        self.swap(i - 1, j);

        true
    }
}

#[cfg(not(debug_assertions))]
macro_rules! mydbg {
    ($($arg:expr),*) => {};
}

macro_rules! echo {
    ($($a:expr),*) => {
          $(println!("{}",$a))*
      }
}

use std::cmp::Ordering::*;
#[allow(unused_imports)]
use std::cmp::{max, min};
use std::collections::{HashMap, HashSet, VecDeque};
use std::io::{stdout, BufWriter, Write};
use std::mem::swap;

fn main() {
    input! {
        n: i32,
        a: i32,
        b: i32,
        mut s: String,
    }
    let mut domestic = 0;
    let mut abroad = 0;
    let mut counter = 0;
    let c:Vec<char> = s.trim().chars().collect();

    for (_, x) in c.iter().enumerate() {
        if *x == 'a' {
            domestic += 1;
            if counter < a + b {
                counter += 1;
                println!("Yes");
            } else {
                println!("No");
            }
        } else if *x == 'b' {
            abroad += 1;
            if counter < a + b && abroad <= b {
                counter += 1;
                println!("Yes");
            } else{
                println!("No");
            }
        } else {
            println!("No");
        }
    }
}