Problem:
Score : 200 points

Problem Statement

You have written N problems to hold programming contests.
The i-th problem will have a score of P_i points if used in a contest.

With these problems, you would like to hold as many contests as possible under the following condition:

A contest has three problems. The first problem has a score not greater than A points, the second has a score between A + 1 and B points (inclusive), and the third has a score not less than B + 1 points.

The same problem should not be used in multiple contests.
At most how many contests can be held?

Constraints

3 \leq N \leq 100

1 \leq P_i \leq 20 (1 \leq i \leq N)

1 \leq A < B < 20

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A B
P_1 P_2 ... P_N

Output

Print the answer.

Sample Input 1

7
5 15
1 10 16 2 7 20 12

Sample Output 1

2

Two contests can be held by putting the first, second, third problems and the fourth, fifth, sixth problems together.

Sample Input 2

8
3 8
5 5 5 10 10 10 15 20

Sample Output 2

0

No contest can be held, because there is no problem with a score of A = 3 or less.

Sample Input 3

3
5 6
5 6 10

Sample Output 3

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

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

#define NDEBUG
#ifdef DEBUG
#include "../cout11.h"
#undef NDEBUG
#endif
#include <cassert>

#ifdef DEBUG
#define debug(fmt, ...) fprintf(stderr, fmt, __VA_ARGS__)
#else
#define debug(fmt, ...)
#endif

typedef long long ll;
typedef long double Double;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<llll> vllll;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<long double> vD;

#define sz(a)  int((a).size())
#define pb  push_back
#define eb  emplace_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define repC3(vari,varj,vark,n)  for(int vari=0;vari<(n)-2;++vari)for(int varj=vari+1;varj<(n)-1;++varj)for(int vark=varj+1;vark<(n);++vark)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
#define found(s,e)  ((s).find(e)!=(s).end())
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))
#define cons make_pair

int solve(int N, int A,int B, vi& a) {
    int p=0,q=0,r=0;
    for (int ai: a){
        if (ai <= A) ++p;
        else if (ai <= B) ++q;
        else ++r;
    }
    return min(min(p,q),r);
}

int main() {
    int N;
    cin >> N;
    int A,B;
    cin >> A >> B;
    vi a(N);
    rep(i,N) cin>>a[i];
    cout << solve(N,A,B,a) << endl;
    return 0;
}