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!

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 ≤ 109) — volume of remaining cola in cans.

The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.

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).

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

【分析】：

we sort the capacities in nonincreasing order and let s = capacity1 + capacity2 if

the answer is no and otherwise the answer is yes。

注意爆int，用LL

【代码】：

#include <bits/stdc++.h>

using namespace std;

typedef  long long ll;

const int N = ;

ll a[N],b[N];

int main()

{

    int n;

    ll sum=;

    scanf("%d",&n);

    for(int i=;i<n;i++)

    {

        scanf("%lld",&a[i]);

        sum+=a[i];

    }

    for(int i=;i<n;i++)

    {

        scanf("%lld",&b[i]);

    }

    sort(b,b+n);

    if(sum<=b[n-]+b[n-])

    {

        printf("YES\n");

    }

    else

    {

        printf("NO\n");

    }

    return ;

}