Problem:
Score : 100 points

Problem Statement

You are given three integers A, B and C.

Determine if there exists an equilateral triangle whose sides have lengths A, B and C.

Constraints

All values in input are integers.

1 \leq A,B,C \leq 100

Input

Input is given from Standard Input in the following format:

A B C

Output

If there exists an equilateral triangle whose sides have lengths A, B and C, print Yes; otherwise, print No.

Sample Input 1

2 2 2

Sample Output 1

Yes

There exists an equilateral triangle whose sides have lengths 2, 2 and 2.

Sample Input 2

3 4 5

Sample Output 2

No

There is no equilateral triangle whose sides have lengths 3, 4 and 5.

Solution in C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <ctype.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>

#define N_MAX (100)
#define P_MAX (100)
#define DP_ARRAY_SIZE (N_MAX * P_MAX / 32 + 1)

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define ABS(a) ((a) < 0 ? -(a) : (a))
#define ABSS(a, b) ((a) > (b) ? (a) - (b) : (b) - (a))

int compare_f(const void *a, const void *b) {
	return *((int64_t *)a) < *((int64_t *)b) ? -1 : 1;
}

int compare_f2(const void *a, const void *b) {
	return *((uint64_t *)a) > *((uint64_t *)b) ? -1 : 1;
}

static size_t comb(const size_t n, const size_t r) {
	size_t result = 1;

	for (size_t i = 0; i < r; i++) {
		result *= n - i;
		result /= i + 1;
	}

	return result;
}

static uint64_t combU64(const uint64_t n, const uint64_t r) {
	uint64_t result = 1;

	for (uint64_t i = 0; i < r; i++) {
		result *= n - i;
		result /= i + 1;
	}

	return result;
}

static uint64_t gcd(uint64_t m, uint64_t n)
{
	uint64_t temp;

	while (m%n != 0) {
		temp = n;
		n = m % n;
		m = temp;
	}

	return n;
}

int main(void) {
	size_t A, B, C;

	scanf("%zu %zu %zu", &A, &B, &C);

	puts((A == B && B == C ? "Yes" : "No"));

	return 0;
}