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 <string.h>
#include <limits.h>
#include <math.h>

int qsort_int(const void *a, const void *b)
{
    // 昇順
    return *(int *) a - *(int *) b;
}

int main(int argc, char *argv[])
{
    int n;
    int l[10];
    int i;
    int sum;

    scanf("%d", &n);

    for (i = 0; i < n; i++) {
        scanf("%d", &l[i]);
    }

    qsort(l, n, sizeof(int), qsort_int);

    sum = 0;
    for (i = 0; i < n - 1; i++) {
        sum += l[i];
    }

    if (l[n - 1] < sum) {
        printf("Yes\n");
    } else {
        printf("No\n");
    }


    return 0;
}