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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
struct P
{
    int x,y,z;
    bool operator<(const P &a)const{
        //if(y!=a.y) return y>a.y;
        return max(x,-x)>max(a.x,-a.x);

    }

};
vector<int> v[1];
//bitset<4001000> b;
int i,n,d,m,k,a,b,c,e;
int o[11];
int l[1];
int j[1];

int dx[10]={0,1,0,-1,1,1,-1,-1},dy[10]={1,0,-1,0,1,-1,1,-1},dz[10]={0,0,0,0,1,-1};
long long mod=1000000007,mod2=1000000009,mod3=2017;
long long x,z,y;
double pi=3.14159265;
P u[1];
stack<int> s;
//set<int> s,s1;

//priority_queue<int> q;
queue<int> q;
//string r1;
char r[1];
//2147483647
//'1'==49;
//'A'==65;
//'a'==97;
//unordered_
//map<int,int> p;
//v.resize(unique(v.begin(),v.end())-v.begin());
/*struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
unordered_map<long long, int, custom_hash> p;//safe_map;
*/
bool as(P a,P b)
{
    //if(a.y!=b.y)
    //return a.y>b.y;
    return a.x<b.x;
}


//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//rng();

int main()
{
    for(int t=0;t<3;t++)
        //puts("@");
        scanf("%d ",&o[t]);
    sort(o+0,o+3);
    if(o[2]==o[0]) puts("Yes");
    else puts("No");

}