Codeforce Round #216 Div2 A

A. Valera and Plates

time limit per test 1 second

memory limit per test 256 megabytes

 

Valera is a lazy student. He has m clean bowls and k clean plates.

Valera has made an eating plan for the next n days. As Valera is lazy, he will eat exactly one dish per day. At that, in order to eat a dish, he needs exactly one clean plate or bowl. We know that Valera can cook only two types of dishes. He can eat dishes of the first type from bowls and dishes of the second type from either bowls or plates.

When Valera finishes eating, he leaves a dirty plate/bowl behind. His life philosophy doesn't let him eat from dirty kitchenware. So sometimes he needs to wash his plate/bowl before eating. Find the minimum number of times Valera will need to wash a plate/bowl, if he acts optimally.

Input

The first line of the input contains three integers n, m, k (1 ≤ n, m, k ≤ 1000) — the number of the planned days, the number of clean bowls and the number of clean plates.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 2). If a_i equals one, then on day i Valera will eat a first type dish. If a_i equals two, then on day i Valera will eat a second type dish.

Output

Print a single integer — the minimum number of times Valera will need to wash a plate/bowl.

Sample test(s)

Input

3 1 1 
1 2 1

Output

1

Input

4 3 1 
1 1 1 1

Output

1

Input

3 1 2 
2 2 2

Output

0

Input

8 2 2 
1 2 1 2 1 2 1 2

Output

4

Note

In the first sample Valera will wash a bowl only on the third day, so the answer is one.

In the second sample, Valera will have the first type of the dish during all four days, and since there are only three bowls, he will wash a bowl exactly once.

In the third sample, Valera will have the second type of dish for all three days, and as they can be eaten from either a plate or a bowl, he will never need to wash a plate/bowl.

#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <queue>

using namespace std;
#define INF 0x7fffffff
#define maxm 1001
#define mod 1000000007
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i = 0; i < (n); i++)
#define re return
#define fi first
#define se second
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define sqrt(x) sqrt(abs(x))
#define y0 y3487465
#define y1 y8687969
#define fill(x,y) memset(x,y,sizeof(x))

typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef double D;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef vector<vi> vvi;

template<class T> T abs(T x) { re x > 0 ? x : -x; }

const int maxn = 100015;

int n, m, t, k,x;
int a[maxn],b[maxn];
char s[maxn];

int main(){
    scanf("%d%d%d", &n, &m, &k);
    int i;
    for (i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    t = 0; x = 0;
    int ans=0;
    for (i = 0; i < n; i++){
        if (a[i] == 1){if (t < m)t++; else ans++;}
        else {
            if (x < k)x++;
            else {
                if (t < m)t++;
                else ans++;
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}