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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<iostream>
#include<vector>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<string.h>
#include<functional>
#include<limits.h>
#include<stdlib.h>
#include<string>
#include<unordered_map>
#include <iomanip>
using namespace std;
 
#define intmax INT_MAX
#define lmax LONG_MAX
#define uintmax UINT_MAX
#define ulmax ULONG_MAX
#define llmax LLONG_MAX
#define ll long long
#define rep(i,a,N) for((i)=(a);(i)<(N);(i)++)
#define rrp(i,N,a) for((i)=(N)-1;(i)>=(a);(i)--)
#define llfor ll i,j,k
#define sc(a) cin>>a
#define pr(a) cout<<a<<endl
#define pY puts("YES")
#define pN puts("NO")
#define py puts("Yes")
#define pn puts("No")
#define pnn printf("\n")
#define sort(a) sort(a.begin(),a.end())
#define push(a,b) (a).push_back(b)
#define llvec vector<vector<ll>>
#define charvec vector<vector<char>>
#define sizeoof(a,b) (a,vector<ll>(b))
#define llpvec vector<pair<ll,ll>>
/*繰り上げ除算*/ll cei(ll x,ll y){ll ans=x/y;if(x%y!=0)ans++;return ans;}
/*最大公約数*/ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
/*最小公倍数*/ll lcm(ll x,ll y){return x/gcd(x,y)*y;}
/*n乗*/ll llpow(ll x,ll n){ll i,ans=1;rep(i,0,n)ans*=x;return ans;}
/*階乗*/ll fact(ll x){ll i,ans=1;rep(i,0,x)ans*=(x-i);return ans;}
/*nCr*/ll ncr(ll n,ll r){return fact(n)/fact(r)/fact(n-r);}
/*nPr*/ll npr(ll n,ll r){return fact(n)/fact(n-r);}
/*primejudge*/bool prime(ll a){if(a<=1)return false;ll i;for(i=2;i*i<=a;i++){if(a%i==0)return false;}return true;}
ll ans=0;llfor;///////////////////////////////////////////////////////////
int main(){
  int A,B,C;
  cin>>A>>B>>C;
  string S="Yes";
  if(A!=B||B!=C||C!=A) S="No";
  cout<<S;
return 0;}