Problem:
Score: 200 points

Problem Statement

Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it.

The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).

Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer.

If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

Constraints

1 \leq N \leq 50000

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If there are values that could be X, the price of the apple pie before tax, print any one of them.

If there are multiple such values, printing any one of them will be accepted.

If no value could be X, print :(.

Sample Input 1

432

Sample Output 1

400

If the apple pie is priced at 400 yen before tax, you have to pay 400 \times 1.08 = 432 yen to buy one.

Otherwise, the amount you have to pay will not be 432 yen.

Sample Input 2

1079

Sample Output 2

:(

There is no possible price before tax for which you have to pay 1079 yen with tax.

Sample Input 3

1001

Sample Output 3

927

If the apple pie is priced 927 yen before tax, by rounding down 927 \times 1.08 = 1001.16, you have to pay 1001 yen.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>

#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  ll_cout(X) printf("%lld\n",X)

typedef long long  ll;
#define rept(x, s, n) for (int x = s; x < n; x++)

#define Yes_RTN s_cout("Yes");return 0
#define No_RTN  s_cout("No");return 0
#define YES_RTN s_cout("YES");return 0
#define NO_RTN  s_cout("NO");return 0


int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);

int s_asorts(const void* a, const void* b);

//昇順
int s_asorts(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	for (i = 0; i < n; i++) {
		i_cin(*(a + i));
	}

	return 0;
}
int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}

long _max(long a, long b) {
	return a > b ? a : b;
}
long _min(long a, long b) {
	return a < b ? a : b;
}
int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}

#pragma endregion



int main(void) {
	int N, tmp;
	
	i_cin(N);

	tmp = round(N / 1.08);
	if ((int)(tmp*1.08) - N)
		s_cout(":(");
	else
		i_cout(tmp);

	return 0;
}