Codeforces Round #446 (Div. 2) A. Greed【模拟】

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 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 ≤ 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.

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".

【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 = 100005;
ll a[N],b[N];
int main()
{
    int n;
    ll sum=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%lld",&a[i]);
        sum+=a[i];
    }
    for(int i=0;i<n;i++)
    {
        scanf("%lld",&b[i]);
    }
    sort(b,b+n);
    if(sum<=b[n-1]+b[n-2])
    {
        printf("YES
");
    }
    else
    {
        printf("NO
");
    }
    return 0;
}