Problem:
Score : 100 points

Problem Statement

You are given three integers, A, B and C.

Among them, two are the same, but the remaining one is different from the rest.

For example, when A=5,B=7,C=5, A and C are the same, but B is different.

Find the one that is different from the rest among the given three integers.

Constraints

-100 \leq A,B,C \leq 100

A, B and C are integers.

The input satisfies the condition in the statement.

Input

Input is given from Standard Input in the following format:

A B C

Output

Among A, B and C, print the integer that is different from the rest.

Sample Input 1

5 7 5

Sample Output 1

7

This is the same case as the one in the statement.

Sample Input 2

1 1 7

Sample Output 2

7

In this case, C is the one we seek.

Sample Input 3

-100 100 100

Sample Output 3

-100

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
int main(void){
  int H,W,bomb;
  scanf("%d %d",&H,&W);
  char masu[H][W];
  for(int i=0;i<H;i++){
    for(int j=0;j<W;j++){
      scanf("%s",&masu[i][j]);
    }
  }


  for(int p=0;p<H;p++){
    for(int q=0;q<W;q++){
      if(masu[p][q]=='.')masu[p][q]='0';
    }
  }


  for(int k=0;k<H;k++){
    for(int l=0;l<W;l++){
      if(masu[k][l]=='#'){
        if(masu[k-1][l-1]!='#') masu[k-1][l-1]+=1;
        if(masu[-1][l]!='#') masu[k-1][l]+=1;
        if(masu[k-1][l+1]!='#')	 masu[k-1][l+1]+=1;
        if(masu[k][l-1]!='#')  masu[k][l-1]+=1;
        if(masu[k][l+1]!='#')  masu[k][l+1]+=1;
        if(masu[k+1][l-1]!='#')	 masu[k+1][l-1]+=1;
        if(masu[k+1][l]!='#') masu[k+1][l]+=1;
        if(masu[k+1][l+1]!='#')	 masu[k+1][l+1]+=1;
      } 
    }
  }

  for(int m=0;m<H;m++){
    for(int n=0;n<W;n++){
      printf("%c",masu[m][n]);
    }
    printf("\n");
  }

  return 0;
}