Problem:
Score : 400 points

Problem Statement

There are N dishes of cuisine placed in front of Takahashi and Aoki.
For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N.

When Takahashi eats Dish i, he earns A_i points of happiness; when Aoki eats Dish i, she earns B_i points of happiness.

Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat.
Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end".

Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_N B_N

Output

Print the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".

Sample Input 1

3
10 10
20 20
30 30

Sample Output 1

20

In this sample, both of them earns 10 points of happiness by eating Dish 1, 20 points by eating Dish 2, and 30 points by eating Dish 3.

In this case, since Takahashi and Aoki have the same "taste", each time they will choose the dish with which they can earn the greatest happiness. Thus, first Takahashi will choose Dish 3, then Aoki will choose Dish 2, and finally Takahashi will choose Dish 1, so the answer is (30 + 10) - 20 = 20.

Sample Input 2

3
20 10
20 20
20 30

Sample Output 2

20

In this sample, Takahashi earns 20 points of happiness by eating any one of the dishes 1, 2 and 3, but Aoki earns 10 points of happiness by eating Dish 1, 20 points by eating Dish 2, and 30 points by eating Dish 3.

In this case, since only Aoki has likes and dislikes, each time they will choose the dish with which Aoki can earn the greatest happiness. Thus, first Takahashi will choose Dish 3, then Aoki will choose Dish 2, and finally Takahashi will choose Dish 1, so the answer is (20 + 20) - 20 = 20.

Sample Input 3

6
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000

Sample Output 3

-2999999997

Note that the answer may not fit into a 32-bit integer.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

//set many funcs template
//Ver.20181228
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<time.h>
#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384

int max(int a,int b){if(a>b){return a;}return b;}
int min(int a,int b){if(a<b){return a;}return b;}
int zt(int a,int b){return max(a,b)-min(a,b);}
int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
int ceil(int a,int b){if(a%b==0){return a/b;}return (a/b)+1;}
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
int nHr(int a,int b){return nCr(a+b-1,b);}
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
int dsum(int x){int r=0;while(x){r+=(x%10);x/=10;}return r;}
int dsumb(int x,int b){int r=0;while(x){r+=(x%b);x/=b;}return r;}
int sankaku(int x){return ((1+x)*x)/2;}
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llnHr(long long a,long long b){return llnCr(a+b-1,b);}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long lldsum(long long x){long long r=0;while(x){r+=(x%10);x/=10;}return r;}
long long lldsumb(long long x,long long b){long long r=0;while(x){r+=(x%b);x/=b;}return r;}
long long llsankaku(long long x){return ((1+x)*x)/2;}
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}
int sortfncsj(const void *a,const void *b){if(*(int *)a>*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int sortfnckj(const void *a,const void *b){if(*(int *)a<*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int llsortfncsj(const void *a,const void *b){if(*(long long *)a>*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int llsortfnckj(const void *a,const void *b){if(*(long long *)a<*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int dbsortfncsj(const void *a,const void *b){if(*(double *)a>*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int dbsortfnckj(const void *a,const void *b){if(*(double *)a<*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int strsortfncsj(const void *a,const void *b){return strcmp((char *)a,(char *)b);}
int strsortfnckj(const void *a,const void *b){return strcmp((char *)b,(char *)a);}
int chsortfncsj(const void *a,const void *b){if(*(char *)a>*(char *)b){return 1;}if(*(char *)a==*(char *)b){return 0;}return -1;}
int chsortfnckj(const void *a,const void *b){if(*(char *)a<*(char *)b){return 1;}if(*(char *)a==*(char *)b){return 0;}return -1;}

void shuffledget(int x[],int n){
    srand(time(0));
    int i,b[524288],p,c;
    for(i=0;i<n;i++){
        b[i]=i;
    }
    for(i=n;i>=1;i--){
        p=rand()%i;
        c=b[i-1];b[i-1]=b[p];b[p]=c;
    }
    for(i=0;i<n;i++){
        scanf("%d",&x[b[i]]);
    }
}

int dx4[4]={1,-1,0,0};
int dy4[4]={0,0,1,-1};
int dx8[8]={-1,-1,-1,0,0,1,1,1};
int dy8[8]={-1,0,1,-1,1,-1,0,1};

int search(int x,int a[],int n){
    int st=0,fi=n-1,te;
    while(st<=fi){
        te=(st+fi)/2;
        if(a[te]<x){st=te+1;}else{fi=te-1;}
    }
    return st;
}

void prarr(int arr[],int n){
  int i;
  for(i=0;i<n;i++){
    if(i){printf(" ");}
    printf("%d",arr[i]);
  }
  printf("\n");
  return;
}

typedef struct{
long long val;
long long av;
long long bv;
}sd;

int sdsortfnc(const void *a,const void *b){
if(((sd*)a)->val > ((sd*)b)->val){return -1;}
if(((sd*)a)->val < ((sd*)b)->val){return 1;}
return 0;
}

int main(void){
    long long i,j,n,m,k,a,b,c,h,w,r=0,l,t;
    sd dat[524288];
    scanf("%lld",&n);
    for(i=0;i<n;i++){
      scanf("%lld%lld",&a,&b);
      dat[i].av=a;
      dat[i].bv=b;
      dat[i].val=a+b;
    }
    qsort(dat,n,sizeof(dat[0]),sdsortfnc);
    a=0;b=0;
    for(i=0;i<n;i++){
      if(i%2==0){a+=dat[i].av;}
      else{b+=dat[i].bv;}
    }
    printf("%lld\n",a-b);
    return 0;
}