CF 1426E

Alice and Bob have decided to play the game "Rock, Paper, Scissors".

The game consists of several rounds, each round is independent of each other. In each round, both players show one of the following things at the same time: rock, paper or scissors. If both players showed the same things then the round outcome is a draw. Otherwise, the following rules applied:

if one player showed rock and the other one showed scissors, then the player who showed rock is considered the winner and the other one is considered the loser;
if one player showed scissors and the other one showed paper, then the player who showed scissors is considered the winner and the other one is considered the loser;
if one player showed paper and the other one showed rock, then the player who showed paper is considered the winner and the other one is considered the loser.

Alice and Bob decided to play exactly

Your task is to find two numbers:

the minimum number of round Alice can win;
the maximum number of rounds Alice can win.

Input

The first line of the input contains one integer

The second line of the input contains three integers

The third line of the input contains three integers

Output

Print two integers: the minimum and the maximum number of rounds Alice can win.

Examples

input

Copy

2
0 1 1
1 1 0

output

Copy

0 1

input

Copy

15
5 5 5
5 5 5

output

Copy

0 15

input

Copy

3
0 0 3
3 0 0

output

Copy

3 3

input

Copy

686
479 178 29
11 145 530

output

Copy

22 334

input

Copy

319
10 53 256
182 103 34

output

Copy

119 226

Note

In the first example, Alice will not win any rounds if she shows scissors and then paper and Bob shows rock and then scissors. In the best outcome, Alice will win one round if she shows paper and then scissors, and Bob shows rock and then scissors.

In the second example, Alice will not win any rounds if Bob shows the same things as Alice each round.

In the third example, Alice always shows paper and Bob always shows rock so Alice will win all three rounds anyway.

题意：Alice 和Bob玩石头剪刀布，两人都确定下来出的石头剪刀布的个数，问Alice最少赢几句最多赢几句。

思路：一看就知道是贪心，问题就是怎么贪，显然最多局数就是把能赢的都赢下来。

　　最少的话，我一开始的思路是让平局的都先平掉，然后把能输的输掉，但是这样有点问题，第二个版本是按照Bob的大小顺序，尽可能让Alice能赢的平掉，再能输的输掉，可惜wa7了。　　

　　最后感觉还是第一种比较对，然后发现只要把第一种略加修改，再平局的时候不去删减Bob的局数就行了，因为有些平局可能因为其他某一个出法太大了而直接用输覆盖掉了而不用平局，所以不去修改Bob的然后就能把Alice赢的可能性降到最小（感觉挺玄学的，有点难讲清楚，需要意会）

下附代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #define ll long long
 5 using namespace std;
 6 int main(){
 7     ll n,s1,j1,b1,s2,j2,b2,ss1,jj1,bb1,ss2,jj2,bb2,flag=1;
 8     scanf("%lld",&n);
 9     scanf("%lld%lld%lld",&s1,&j1,&b1);
10     ss1=s1,jj1=j1,bb1=b1;
11     scanf("%lld%lld%lld",&s2,&j2,&b2);
12     ss2=s2,jj2=j2,bb2=b2;
13     ss1-=ss2,jj1-=jj2,bb1-=bb2;
14     if (ss1>0) ss1-=bb2;
15     if (jj1>0) jj1-=ss2;
16     if (bb1>0) bb1-=jj2;
17     ss1=max(0ll,ss1);
18     bb1=max(0ll,bb1);
19     jj1=max(0ll,jj1);
20     printf("%lld %lld",min(ss1,jj2)+min(jj1,bb2)+min(bb1,ss2),min(s1,j2)+min(j1,b2)+min(b1,s2));
21     return 0;
22 }

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