Problem:
Score : 100 points

Problem Statement

Snuke is buying a bicycle.
The bicycle of his choice does not come with a bell, so he has to buy one separately.

He has very high awareness of safety, and decides to buy two bells, one for each hand.

The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively.
Find the minimum total price of two different bells.

Constraints

1 \leq a,b,c \leq 10000

a, b and c are integers.

Input

Input is given from Standard Input in the following format:

a b c

Output

Print the minimum total price of two different bells.

Sample Input 1

700 600 780

Sample Output 1

1300

Buying a 700-yen bell and a 600-yen bell costs 1300 yen.

Buying a 700-yen bell and a 780-yen bell costs 1480 yen.

Buying a 600-yen bell and a 780-yen bell costs 1380 yen.

The minimum among these is 1300 yen.

Sample Input 2

10000 10000 10000

Sample Output 2

20000

Buying any two bells costs 20000 yen.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(V) (V).begin(),(V).end()
#define SORT(V) sort(ALL(V)) //小さい方からソート
#define REV(V) reverse(ALL(V)) //リバース
#define RSORT(V) SORT(V);REV(V) //大きい方からソート
#define NPN(V) next_permutation(ALL(V)) //順列
#define pb(n) push_back(n)
#define endl '\n'
#define Endl '\n'
#define DUMP(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define Yay(n) cout << ((n) ? "Yay!": ":(") << endl
#define RAPID cin.tie(0);ios::sync_with_stdio(false)
#define IN(n) cin >> n
#define IN2(a,b) cin >> a >> b
#define IN3(a,b,c) cin >> a >> b >> c 
#define IN4(a,b,c,d) cin >> a >> b >> c >> d
#define VIN(V) for(int i = 0; i < (V).size(); i++) {cin >> (V).at(i);}
#define OUT(n) cout << n << endl
#define VOUT(V) REP(i, (V).size()) {cout << (V)[i] << endl;}
#define VOUT2(V) REP(i, (V).size()) {cout << (V)[i] << " ";} cout<<endl;

// 型マクロ定義
#define int long long
#define P pair<ll,ll>
#define Vi vector<ll>
#define Vd vector<double>
#define Vs vector<string>
#define Vc vector<char>
#define M map<ll,ll>
#define S set<ll>
#define PQ priority_queue<ll>
#define PQG priority_queue<ll,V,greater<ll>
//

const int MOD = 1000000007;
const int INF = 1061109567;
const double EPS = 1e-10;
const double PI  = acos(-1.0);

int gcd(int a,int b){return b?gcd(b,a%b):a;} //最小公倍数

// デフォルト変数定義
int n,m,a,b,c,d,x,y,z;
string s,t;
//

signed main() {
  RAPID;
  IN3(a,b,c);
  OUT(min({a+b,b+c,c+a}));
}