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Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai  ≤  bi).

Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!

Input

The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 10^9) — volume of remaining cola in cans.

The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 10^9) — capacities of the cans.

 Output

Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).

You can print each letter in any case (upper or lower).

Examples

Input

2

3  5

3  6

Output

 YES

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Input

3

6 8 9

6 10 12

Output

 NO

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Input

5

0 0 5 0 0

1 1 8 10 5

Output

 YES

---

Input

4

4 1 0 3

5 2 2 3

Output

 YES

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Note

In the first sample, there are already 2 cans, so the answer is "YES".

题意：

其实这个题意是比较简单的，要求第一行给出一个整数n表示n罐可乐，第二行是n个数来表示n个罐中剩余的可乐的体积，第三行是n个数来表示这n个罐的容量。然后来判断这些剩余的所有可乐能否全部装到n个罐中任选的两个罐中。

思路：

计算出所有剩余可乐的体积，然后与n个罐中体积最大的两个罐来比较就可以了。

代码：

#include
   
    #include
    
     using namespace std;int main(){   long long i,n,x,y,sum=0;   long long max1=0,max2=0;  //max1来存放最大的可乐罐的体积，max2来存放第二大的可乐罐体积   scanf("%lld",&n);   for(i=0;i
     
      =max1)     {       max2=max1;       max1=y;     }     else if(y>max2)     {       max2=y;     }   }   if(sum <= max1+max2)     printf("Yes\n");   else      printf("No\n");      return 0;}
     
    
   

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