It is easy to see that for every fraction in the form  (k > 0), we can always find two positive integers x and y, x ≥ y, such that: 

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Now our question is: can you write a program that counts how many such pairs of x and y there are for any given k?

Input

Input contains no more than 100 lines, each giving a value of k (0 < k ≤ 10000).

Output

For each k, output the number of corresponding (x, y) pairs, followed by a sorted list of the values of x and y, as shown in the sample output.

Sample Input

212

Sample Output

21/2 = 1/6 + 1/31/2 = 1/4 + 1/481/12 = 1/156 + 1/131/12 = 1/84 + 1/141/12 = 1/60 + 1/151/12 = 1/48 + 1/161/12 = 1/36 + 1/181/12 = 1/30 + 1/201/12 = 1/28 + 1/211/12 = 1/24 + 1/24

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4  5 using namespace std; 6  7 char ch[1000][50]; 8 int main () { 9     int n;10     while (cin >> n) {11         int pos = 0;12         for (int i = n + 1;i <= 2 * n;i++) {13             if (i * n % (i - n) == 0) {14                 sprintf(ch[pos++],"1/%d = 1/%d + 1/%d\n",n,i * n / (i - n),i);15             }16         }17         cout << pos << endl;18         for (int i = 0;i < pos;i++) {19             cout << ch[i];20         }21     }22 }

View Code

UVA - 10976