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)

/*
ID: seekho
PROG: friday
LANG: C++14                 
*/
/* LANG can be C++11 or C++14 for those more recent releases */

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update

using namespace std;
using namespace __gnu_pbds;

//find_by_order gives kth-largest element
//order_of_key gives the number of items in the set that are strictly smaller!!

// OT.find_by_order(k-1)
// 
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define db4(x, y, z, w) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << "," << #w << "=" << w << endl
#define ll long long
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define mp make_pair
#define X first
#define Y second
#define sz(x) (int)((x).size())
#define pii pair<int,int>
#define MOD (ll)(998244353)
#define rep(i, n)    for(int i = 0; i < (n); ++i)
#define repA(i, a, n)  for(int i = a; i <= (n); ++i)
#define repD(i, a, n)  for(int i = a; i >= (n); --i)
#define trav(a, x) for(auto& a : x)
typedef vector<int> vi;
#define inf (ll)(1e18)
#define double long double
#define int long long

// #define N 50005


int32_t main() {

    ios::sync_with_stdio(false);
    cin.tie(0); 
    cout.tie(0);

    int n,a,b;
    cin >> n >> a >> b;



    vector<int> ai[3];
    rep(i,n) {
        int aa;
        cin >> aa;
        // cin >> a[i];
        if(aa <= a) ai[0].pb(1);
        else if(aa > a and aa <= b) ai[1].pb(1);
        else ai[2].pb(1);
    }

    int ans = min({sz(ai[0]), sz(ai[1]), sz(ai[2])});

    cout << ans << "\n";


}