Contest Duration: - (local time) (120 minutes)
C - Different Strokes /

Time Limit: 2 sec / Memory Limit: 1024 MB

問題文

このとき、「最終的に高橋くんが得る幸福度の総和」から「最終的に青木さんが得る幸福度の総和」を引いた値はいくつになるでしょうか？

制約

• 1 \leq N \leq 10^5
• 1 \leq A_i \leq 10^9
• 1 \leq B_i \leq 10^9
• 入力される値はすべて整数である。

入力

N
A_1 B_1
:
A_N B_N


出力

「最終的に高橋くんが得る幸福度の総和」から「最終的に青木さんが得る幸福度の総和」を引いた値を出力せよ。

入力例 1

3
10 10
20 20
30 30


出力例 1

20


この例では、二人のどちらも、料理 1 を食べると 10 ポイント、料理 2 を食べると 20 ポイント、料理 3 を食べると 30 ポイントの幸福度を得ます。

この場合、高橋くんと青木さんの「好み」が一致しているため、彼らは毎回残っている料理のうち最も幸福度を多く得られる料理を選びます。よって、最初に高橋くんは料理 3 を選び、次に青木さんは料理 2 を選び、最後に高橋くんが料理 1 を選ぶため、答えは (30 + 10) - 20 = 20 です。

入力例 2

3
20 10
20 20
20 30


出力例 2

20


この例では、高橋くんは料理 1,2,3 のいずれを食べても 20 ポイントの幸福度を得ますが、青木さんは料理 1 を食べると 10 ポイント、料理 2 を食べると 20 ポイント、料理 3 を食べると 30 ポイントの幸福度を得ます。

入力例 3

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


出力例 3

-2999999997


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.