Consider the following function f defined for any natural number n:

f(n) is the number obtained by summing up the squares of the digits of n in decimal (or base-ten).

  If n = 19, for example, then f(19) = 82 because 1^2 + 9^2 = 82.

  Repeatedly applying this function f, some natural numbers eventually become 1. Such numbers are called happy numbers. For example, 19 is a happy number, because repeatedly applying function to 19 results in:

  f(19) = 1^2 + 9^2 = 82

  f(82) = 8^2 + 2^2 = 68

  f(68) = 6^2 + 8^2 = 100

  f(100) = 12 + 02 + 02 = 1

  However, not all natural numbers are happy. You could try 5 and you will see that 5 is not a happy number. If n is not a happy number, it has been proved by mathematicians that repeatedly applying function f to n reaches the following cycle:

4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4.

  Write a program that decides if a given natural number n is a happy number or not.

Input

The input file contains several test cases, each of them consists of a single line that contains an integer, n (1 ≤ n ≤ 1, 000, 000, 000)

Output

For each test case, print exactly one line. If the given number n is a happy number, print out ‘HAPPY’; otherwise, print out ‘UNHAPPY’.

Sample Input

19

5

Sample Output

HAPPY

UNHAPPY

問題簡述：（略）

  某一個正整數n，對其各位數字分別平方再求和得到一個新數，重複同樣的計算，最終和變成1，則稱n為快樂數；如果出現迴圈變不成1則不是快樂數。

程式說明：

  使用set來判重複是一個好做法。

  函式ishn()是CV來的，其中包含統計步數的邏輯，參見參考連結。

題記：（略）

AC的C++語言程式如下：

/* UVALive8091 Happy Number */

#include <bits/stdc++.h>

using namespace std;

int ishn(int n) {
    set<int> s;
    int step = 1;
    while (n != 1) {
        step++;

        int sum = 0;
        while (n) {
            int d = n % 10;
            sum += d * d;
            n /= 10;
        }
        n = sum;
        if (s.count(n)) break;
        else s.insert(n);
    }
    return n == 1 ? step : 0;
}

int main()
{
    int n;

    while(~scanf("%d", &n))
        printf("%s\n", ishn(n) ? "HAPPY" : "UNHAPPY");

    return 0;
}