If at first you don't succeed...（CodeForces 991A）（简单容斥）

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本文通过一组学生考试后是否庆祝及选择庆祝地点的数据，提出了一个有趣的问题：如何根据已知的庆祝人数推断未通过考试的学生人数。文章还提供了一段代码来解决这个问题。

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Description

Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.

Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by AA students, BeaverKing — by BB students and CC students visited both restaurants. Vasya also knows that there are NN students in his group.

Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?

Input

The first line contains four integers — AA, BB, CC and NN (0≤A,B,C,N≤1000≤A,B,C,N≤100).

Output

If a distribution of NN students exists in which AA students visited BugDonalds, BB — BeaverKing, CC — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.

If such a distribution does not exist and Vasya made a mistake while determining the numbers AA, BB, CC or NN (as in samples 2 and 3), output −1−1.

Sample Input

Input

10 10 5 20

Output

Input

2 2 0 4

Output

-1

Input

2 2 2 1

Output

-1

Hint

The first sample describes following situation: 55 only visited BugDonalds, 55 students only visited BeaverKing, 55 visited both of them and 55 students (including Vasya) didn't pass the exam.

In the second sample 22 students only visited BugDonalds and 22 only visited BeaverKing, but that means all 44 students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.

The third sample describes a situation where 22 students visited BugDonalds but the group has only 11 which makes it clearly impossible.

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<iomanip>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<utility>
#include<list>
#include<algorithm>
#include <ctime>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define memset(a,v)  memset(a,v,sizeof(a))
#define X (sqrt(5)+1)/2.0
#define maxn 320007
#define N 200005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e-8
#define ll long long
using namespace std;



int main()
{
    int a,b,c,m;
    cin>>a>>b>>c>>m;
    if(a>m||b>m||c>m||c>a||c>b)
        cout<<"-1"<<endl;
    else if(a+b-c>=m)
        cout<<"-1"<<endl;
    else
        cout<<m-(a+b-c)<<endl;
    return 0;
}