Problem:
Score : 200 points

Problem Statement

Determine if an N-sided polygon (not necessarily convex) with sides of length L_1, L_2, ..., L_N can be drawn in a two-dimensional plane.

You can use the following theorem:

Theorem: an N-sided polygon satisfying the condition can be drawn if and only if the longest side is strictly shorter than the sum of the lengths of the other N-1 sides.

Constraints

All values in input are integers.

3 \leq N \leq 10

1 \leq L_i \leq 100

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 ... L_N

Output

If an N-sided polygon satisfying the condition can be drawn, print Yes; otherwise, print No.

Sample Input 1

4
3 8 5 1

Sample Output 1

Yes

Since 8 < 9 = 3 + 5 + 1, it follows from the theorem that such a polygon can be drawn on a plane.

Sample Input 2

4
3 8 4 1

Sample Output 2

No

Since 8 \geq 8 = 3 + 4 + 1, it follows from the theorem that such a polygon cannot be drawn on a plane.

Sample Input 3

10
1 8 10 5 8 12 34 100 11 3

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<bits/stdc++.h>
typedef long long int ll;
#define REP(i, n) for(int i = 0; i < (n); i++)
#define FOR_IN(i, a, b) for(int i = (a); i < (b); i++)
#define BETWEEN(x, a, b) ((x) >= (a) && (x) <= (b))
#define BIT(b, i) (((b) >> (i)) & 1)
#define LOG_F 1
#define LOG(...) if(LOG_F) fprintf(stderr, __VA_ARGS__)

using namespace std;

ll pow(int x, int n){
  return n == 0 ? 1 : x * pow(x, n - 1);
}


/*
n
l1 ... ln
 */
int n,l[10];

int main(){
  cin>>n;
  REP(i,n)
    cin>>l[i];

  sort(l,l+n);
  int s=0;
  REP(i,n-1)
    s+=l[i];

  if(l[n-1] < s)
    cout<<"Yes\n";
  else
    cout<<"No\n";

  return 0;
}