Problem:
Score : 300 points

Problem Statement

Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.

Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Constraints

1 \leq K \leq 200

K is an integer.

Input

Input is given from Standard Input in the following format:

K

Output

Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.

Sample Input 1

2

Sample Output 1

9

\gcd(1,1,1)+\gcd(1,1,2)+\gcd(1,2,1)+\gcd(1,2,2)
+\gcd(2,1,1)+\gcd(2,1,2)+\gcd(2,2,1)+\gcd(2,2,2)
=1+1+1+1+1+1+1+2=9

Thus, the answer is 9.

Sample Input 2

200

Sample Output 2

10813692

Solution in JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)


process.stdin.resume();
process.stdin.setEncoding('utf8');

var lines = []
var reader = require('readline').createInterface({
  input: process.stdin,
  output: process.stdout
});

reader.on('line', (line) => {
  lines.push(line);
});

reader.on('close', () => {
 var k = Number(lines[0]);
 var one = 1;
 var two = 1;
 var three = 1;
 var ans = 0;
 var n = 0;
 
 while(n !== k){
   ans += gcd(one,two,three);
   if(one == two && two == three){
     three = three + 1;
   }else if((one == two && two !== three) && one < three){
     two = two + 1;
     three = three - 1;
   }else if((one == three && one !== two) && one < two){
     three = three + 1;
   }else if((two == three && one != two) && (one < two)){
     one = one + 1;
   }else if((two == three && one != two) && (one > two)){
     three = three + 1;
   }else if((one == three && one !== two) && (one > two)){
     two = two + 1;
     three = three - 1;
   }else if((one == two && two !== three) && (one > three)){
     three = three + 1;
   }
   
   n++;
 }
 
 console.log(ans);
 
function gcd() {
     var f = (a, b) => b ? f(b, a % b) : a
     var ans = arguments[0]
     for (var i = 1; i < arguments.length; i++) {
         ans = f(ans, arguments[i]); 
     }
     return ans;
 
 

}
 
  
 
});