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".
#include <bits/stdc++.h>
using namespace std;
const int N = 300010;
const int M = 1000010;
const int MOD = 1000000007;
const int INF = 2e9;
const double eps = 1e-8;
const double PI = acos(-1.0);
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int x, int y) {
  x /= gcd(x, y);
  return x * y;
}
int Min(int a, int b) { return a < b ? a : b; }
int Max(int a, int b) { return a > b ? a : b; }
bool prime[M + 5], isprime[M + 5];
vector<int> v;
int n, m, q, cs, ar[123456];
void sieve() {
  v.push_back(2);
  prime[1] = 1;
  int limit = (int)sqrt(M);
  for (int i = 4; i <= M; i += 2) prime[i] = 1;
  for (int i = 3; i <= M; i += 2) {
    if (!prime[i]) {
      v.push_back(i);
      if (i <= limit) {
        for (int j = i * i; j <= M; j += 2 * i) prime[j] = 1;
      }
    }
  }
}
int br[123456];
int main() {
  long long n, tot = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    tot += ar[i];
  }
  for (int i = 0; i < n; i++) cin >> br[i];
  sort(br, br + n);
  long long a = br[n - 1], b = br[n - 2];
  if (tot <= a + b)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
