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>

int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

int main(){
    int N, L[15];
    int sum = 0;
    scanf("%d", &N);
    for (int c = 0; c < N;c++){
        scanf("%d", &L[c]);
    }
     qsort(L, N, sizeof(int), compare_int);
     
        for (int d = 0; d < N-1;d++){
         sum += L[d];
                }
                if (sum > L[N - 1] ){
                        printf("Yes");
                    }else{
                        printf("No");
                    }

                    return 0;
}