codeforces 892 A题 Greed

A. Greed

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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, ..., a_n (0 ≤ a_i ≤ 10⁹) — volume of remaining cola in cans.

The third line contains n space-separated integers that b₁, b₂, ..., b_n (a_i ≤ b_i ≤ 10⁹) — 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

Input

3
6 8 9
6 10 12

Output

NO

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

Note

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

题意：有 n 瓶可乐 ， 然后每瓶可乐都有剩余的可乐 和 不同的容量 ， 问能不能把所有剩余的可乐倒在 两个可乐瓶子里。

思路：累加可乐剩余的体积 ， 找出 最大的两个可乐的 容量和  ， 比较

#include <iostream>
#include <algorithm>

using namespace std ; 

#define maxn 101000
#define LL long long 
 
int main(){
    int n ; 
    LL num[maxn] ; 
    LL sum[maxn] ; 
    while(~scanf("%d" , &n)){
        
    LL summ = 0 ;  
        for(int i=1 ; i<= n ; i++){
            scanf("%lld" , &num[i]) ; 
            summ += num[i] ; 
        } 
        
        for(int i=1 ; i<=n ; i++){
            scanf("%lld" , &sum[i]) ; 
        } 
    
        sort(sum+1 , sum+1+n ) ; 
        if(summ <= sum[n] + sum[n-1]) {
            printf("YES\n") ; 
        } else {
            printf("NO\n") ; 
        }
    
    } 
    return 0 ; 
}