Problem: 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".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
const T& max(const T& a, const T& b, const T& c) {
  return max(a, max(b, c));
}
template <class T>
const T& min(const T& a, const T& b, const T& c) {
  return min(a, min(b, c));
}
void sc(int& a) { scanf("%d", &a); }
void sc(long long& a) { scanf("%lld", &a); }
void sc(int& a, int& b) {
  sc(a);
  sc(b);
}
void sc(long long& a, long long& b) {
  sc(a);
  sc(b);
}
void sc(int& a, int& b, int& c) {
  sc(a, b);
  sc(c);
}
void sc(long long& a, long long& b, long long& c) {
  sc(a, b);
  sc(c);
}
void prl(int a) { printf("%d\n", a); }
void prl(long long a) { printf("%lld\n", a); }
void prl() { printf("\n"); }
void prs(int a) { printf("%d ", a); }
void prs(long long a) { printf("%lld ", a); }
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long poww(long long a, long long b) {
  if (b == 0) return 1;
  long long tmp = poww(a, b / 2);
  return (b & 1 ? a * tmp * tmp : tmp * tmp);
}
long long sumOfDigs(string s) {
  long long sum = 0;
  for (int i = 0; i < s.length(); i++) sum += s[i] - '0';
  return sum;
}
long long sumOfDigs(long long n) {
  return (n < 10 ? n : n % 10 + sumOfDigs(n / 10));
}
string itos(long long i) {
  string s = "";
  while (i) {
    s += char(i % 10 + '0');
    i /= 10;
  }
  reverse(s.begin(), s.end());
  return s;
}
long long stoi(string& s) {
  long long tot = 0;
  for (int i = (int)s.length() - 1, j = 1; i >= 0; i--, j *= 10) {
    tot += j * (s[i] + '0');
  }
  return tot;
}
int months[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
using namespace std;
long long mod = 1e9 + 7;
void tt() { freopen("test.txt", "r", stdin); }
long long modpower(long long x, long long y, long long p) {
  x %= mod;
  if (!y) return 1;
  long long res = 1;
  if (y & 1) {
    res *= x;
    res %= p;
  }
  long long z = modpower(x, y / 2, p);
  z %= p;
  z *= z;
  z %= mod;
  res *= z;
  res %= p;
  return res;
}
struct HASH {
  size_t operator()(const pair<int, int>& x) const {
    size_t ans = 0;
    for (int i = 0; i < x.first; i++) ans += x.second;
    return ans;
  }
};
const int MAX = 1e6 + 10;
int r[MAX], c[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  long long tot = 0;
  for (int i = 1; i <= n; i++) cin >> r[i];
  for (int i = 1; i <= n; i++) tot += r[i];
  for (int i = 1; i <= n; i++) cin >> c[i];
  sort(c + 1, c + n + 1);
  long long x = c[n] + c[n - 1];
  cout << (x >= tot ? "YES" : "NO") << '\n';
  return 0;
}