Problem:
Score : 200 points

Problem Statement

We have A cards, each of which has an integer 1 written on it. Similarly, we also have B cards with 0s and C cards with -1s.

We will pick up K among these cards. What is the maximum possible sum of the numbers written on the cards chosen?

Constraints

All values in input are integers.

0 \leq A, B, C

1 \leq K \leq A + B + C \leq 2 \times 10^9

Input

Input is given from Standard Input in the following format:

A B C K

Output

Print the maximum possible sum of the numbers written on the cards chosen.

Sample Input 1

2 1 1 3

Sample Output 1

2

Consider picking up two cards with 1s and one card with a 0.
In this case, the sum of the numbers written on the cards is 2, which is the maximum possible value.

Sample Input 2

1 2 3 4

Sample Output 2

0

Sample Input 3

2000000000 0 0 2000000000

Sample Output 3

2000000000

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define SIZE 100
#define rep(i, N) for (i = 0; i < N; i++) //制御変数iを用いてN回転
#define array(N, t) ((t)*)calloc(N, sizeof(t)) //t型N要素の配列を動的確保後0クリア(freeを忘れずに)
#define zero(a); {int iter; rep(iter, sizeof(a)/sizeof((a)[0])) (a)[iter]=0;} //1次元配列aのゼロ初期化?
typedef long long ll;
void printa_int(int *a, int size); //int型の1次元配列aを出力(for debug)

int main() {
    ll A, B, C, K;
    scanf("%lld %lld %lld %lld", &A, &B, &C, &K);
    if (A >= K) printf("%lld\n", K);
    else if ((A + B) >= K) printf("%lld\n", A);
    else printf("%lld\n", 2 * A + B - K);
    return 0;
}

void printa_int(int *a, int size) {
    int i;
    for (i = 0; i < size; i++)
        printf("%d, ", a[i]);
    printf("\n");
}