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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
 
 
void swap (int *x, int *y) {
  int temp;    
 
  temp = *x;
  *x = *y;
  *y = temp;
}
 
int gcd(int a,int b){
  if(a<b){
    swap(&a,&b);
  }
 
  if(b==0){
    return a;
  }
 
  int r=a%b;
  while(r!=0){
    a = b;
    b = r;
    r = (a%b);
  }
 
  return b;
}
 
/* クイックソート */
int dn(const void*a,const void*b){return*(int*)b-*(int*)a;}
 
 
int modpow(int a,int n,int mod){
  int ans=1;
  while(n>0){
    if(n & 1){
      ans=ans*a%mod;
    }
    a=a*a%mod;
    n/=2;
  }
  return ans;
}
 
int max(int a,int b){
  if(a<b){
    return b;
  }
  else{
    return a;
  }
}
 
int min(int a,int b){
  if(a<b){
    return a;
  }
  else{
    return b;
  }
}
 
void chmax(int *a,int b){
  if(*a < b){
    *a = b;
  }
 
  return;
}
 
void chmin(int *a,int b){
  if(*a > b){
    *a =b;
  }
 
  return;
}
 

int main(void){
  int n;
  scanf("%d",&n);
  int l[n];
  int max=0;
  int sum=0;
  for(int i=0;i<n;i++){
    scanf("%d",&l[i]);
    chmax(&max, l[i]);
    sum+=l[i];
  }
  sum-=max;
  if(max<sum){
    printf("Yes\n");
    return 0;
  }else{
    printf("No\n");
    return 0;
  }
}