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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

var (
	finv, inv, fac [MAX]int
)

func COM(n, k int) int {
	MOD := int(MOD)
	if n < k {
		return 0
	}
	if n < 0 || k < 0 {
		return 0
	}
	return fac[n] * (finv[k] * finv[n-k] % MOD) % MOD
}

func init() {
	MOD := int(MOD)
	fac[0] = 1
	fac[1] = 1
	finv[0] = 1
	finv[1] = 1
	inv[1] = 1
	for i := 2; i < MAX; i++ {
		fac[i] = fac[i-1] * i % MOD
		inv[i] = MOD - inv[MOD%i]*(MOD/i)%MOD
		finv[i] = finv[i-1] * inv[i] % MOD
	}
	//sc.Split(bufio.ScanLines)
	sc.Split(bufio.ScanWords)
	sc.Buffer([]byte{}, 500100)
}

const (
	MAX = 510000
)

var MOD int64 = 1000000007

func ModAdd(a, b int64) int64 {
	return (a + b) % MOD
}

func ModMul(a, b int64) int64 {
	ans := a * b % MOD
	if ans < 0 {
		ans += MOD
	}
	return ans
}

func ModSub(a, b int64) int64 {
	r := (a%MOD - b%MOD) % MOD
	if r < 0 {
		r += MOD
	}
	return r
}

func ModDiv(a, b int64) int64 {
	return a % MOD * ModInv(b) % MOD
}

func ModPow(a, n int64) int64 {
	var r int64 = 1
	for n > 0 {
		if n&1 == 1 {
			r = r * a % MOD
		}
		a = a * a % MOD
		n >>= 1
	}
	return r
}

func ModInv(a int64) int64 {
	var b, u, v int64 = MOD, 1, 0
	for b > 0 {
		t := a / b
		a -= t * b
		a, b = b, a
		u -= t * v
		u, v = v, u
	}
	u %= MOD
	if u < 0 {
		u += MOD
	}
	return u
}

type Q struct {
	value int
	index int
	A, B  int
}

type QS []*Q

func (p QS) Len() int           { return len(p) }
func (p QS) Less(i, j int) bool { return p[i].value > p[j].value }
func (pq QS) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}

func (pq *QS) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Q)
	item.index = n
	*pq = append(*pq, item)
}

func (pq *QS) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	old[n-1] = nil  // avoid memory leak
	item.index = -1 // for safety
	*pq = old[0 : n-1]
	return item
}

func GCD(a, b int64) int64 {
	if b == 0 {
		return a
	}
	return GCD(b, a%b)
}

func LCM(a, b int64) int64 {
	return a * b / GCD(a, b)
}

type UnionFind struct {
	par  []int
	size []int
}

func NewUnionFind(size int) *UnionFind {
	return new(UnionFind).init(size)
}

func (uf *UnionFind) init(size int) *UnionFind {
	uf.par = make([]int, size)
	uf.size = make([]int, size)

	for i := 0; i < size; i++ {
		uf.par[i] = i
		uf.size[i] = 1
	}
	return uf
}

func (uf *UnionFind) Unite(p int, q int) {
	rp := uf.Root(p)
	rq := uf.Root(q)
	if rp == rq {
		return
	}
	if uf.size[rp] < uf.size[rq] {
		uf.par[rp] = uf.par[rq]
		uf.size[rq] += uf.size[rp]
	} else {
		uf.par[rq] = uf.par[rp]
		uf.size[rp] += uf.size[rq]
	}
}

func (uf *UnionFind) Root(p int) int {
	for uf.par[p] == p {
		return p
	}
	uf.par[p] = uf.Root(uf.par[p])
	return uf.par[p]
}

func (uf *UnionFind) Size(p int) int {
	return uf.size[uf.Root(p)]
}

func (uf *UnionFind) Same(p int, q int) bool {
	return uf.Root(p) == uf.Root(q)
}

var sc = bufio.NewScanner(os.Stdin)

func getInt() int {
	n, _ := strconv.Atoi(getStr())
	return n
}

func getStr() string {
	sc.Scan()
	return sc.Text()
}

func factor(result map[int]int, num, pnum int) map[int]int {
	if pnum*pnum > num {
		if num != 1 {
			result[num] += 1
		}
		return result
	}

	if num%pnum == 0 {
		num /= pnum
		result[pnum]++
	} else {
		pnum++
	}
	return factor(result, num, pnum)
}

func main() {
	N := getInt()
	A := make([]int, N)
	B := make([]int, N)
	C := make([]int, N)
	for i := 0; i < N; i++ {
		A[i] = getInt()
		B[i] = getInt()
		C[i] = A[i] + B[i]
	}
	sort.Sort(sort.Reverse(sort.IntSlice(C)))
	l := N / 2
	if N%2 != 0 {
		l++
	}
	var ans int64
	for i := 0; i < l; i++ {
		ans += int64(C[2*i])
	}
	for i := 0; i < N; i++ {
		ans -= int64(B[i])
	}
	//fmt.Println(A, B, C)
	fmt.Println(ans)
}

type IntHeap []int

func (h IntHeap) Len() int           { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h IntHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IntHeap) Push(x interface{}) {
	// Push and Pop use pointer receivers because they modify the slice's length,
	// not just its contents.
	*h = append(*h, x.(int))
}

func (h *IntHeap) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

func Reverse(s string) string {
	runes := []rune(s)
	for i, j := 0, len(runes)-1; i < j; i, j = i+1, j-1 {
		runes[i], runes[j] = runes[j], runes[i]
	}
	return string(runes)
}