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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define ll long long
#define rep(i,n) for(ll i=0;i<(n);i++)
#define max(p,q) ((p)>(q)?(p):(q))
#define min(p,q) ((p)<(q)?(p):(q))
#define chmax(a,b) ((a)=(a)>(b)?(a):(b))
#define chmin(a,b) ((a)=(a)<(b)?(a):(b))
#define abs(p) ((p)>=(0)?(p):(-(p)))
#define swap(a,b) do{ll w=(a);(a)=(b);(b)=w;}while(0)
#define swapd(a,b) do{double w=(a);(a)=(b);(b)=w}while(0)
#define in(a) scanf("%lld", &(a))
#define ind(a) scanf("%lf", &(a))
#define put(a) printf("%lld\n", (a))
#define putd(a) printf("%.15f\n", (a));
//def   puts(a) printf("%s\n", a) 文字はこっち
//your code here!


int main(void){
    ll N, L[11], MAX=0, sum=0;
    in(N);
    rep(i, N){
        in(L[i]);
        chmax(MAX, L[i]);
        sum += L[i];
    }
    puts(MAX<sum-MAX?"Yes":"No");
    return 0;
}