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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

// 型名の省略
typedef unsigned int uint;
typedef long long lint;
typedef unsigned long long ulint;
typedef pair<int, int> P;
typedef pair<lint, lint> Plint;
 
const int MOD = 1e9+7;
const int INF = 1<<18;
const lint INFLINT = 1LL<<60;
const double PI=acos(-1.0);
 
// cinとcoutの高速化
#define endl "\n"
struct cinacc{cinacc(){cin.tie(0);ios::sync_with_stdio(false);}}cinaccc;
struct coutacc{coutacc(){cout.setf(ios::fixed);cout.precision(12);}}coutaccc;
 
// 繰り返し
#define rep(i,n) for(lint i=0;i<(lint)(n);i++)
#define reps(i,n) for(lint i=0;i<=(lint)(n);i++)
#define rrep(i,n) for(lint i=(lint)(n)-1;i>0;i--)
#define rreps(i,n) for(lint i=(lint)(n)-1;i>=0;i--)

#define pb push_back

#define ALL(s) (s.begin(), s.end())
#define SZ(s) (s.size())

template<class T>bool chmax(T&a,const T&b){if(a<b){a=b;return true;}return false;}
template<class T>bool chmin(T&a,const T&b){if(b<a){a=b;return true;}return false;}

// Union-Find-Tree
class UnionFind {
	vector<int> par;
	UnionFind(int n):par(n){rep(i,n)par[i]=i;}
	int root(int x){if(par[x]==x)return x;return par[x]=root(par[x]);}
	void unite(int x,int y){int rx=root(x);int ry=root(y);if(rx==ry)return;par[rx]=ry;}
	bool same(int x,int y){return root(x)==root(y);}
};

// Final Weapon!
// #define int lint

int main() {
    int n, a, b;
    string s;
    cin >> n >> a >> b >> s;
    a--; b--;
    
    int cnt = 0, cnt2 = 0;
    rep(i, n) {
        if(s[i] == 'a') {
            if(a + b > cnt) {
                cout << "Yes" << endl;
                cnt++;
            } else {
                cout << "No" << endl;
            }
        } else if(s[i] == 'b') {
            if(a + b > cnt && b >= cnt2) {
                cout << "Yes" << endl;
                cnt2++;
            } else {
                cout << "No" << endl;
            }
        } else if(s[i] == 'c') {
            cout << "No" << endl;
        }
    }
    
    return 0;
}