Problem:
Score : 100 points

Problem Statement

There are three airports A, B and C, and flights between each pair of airports in both directions.

A one-way flight between airports A and B takes P hours, a one-way flight between airports B and C takes Q hours, and a one-way flight between airports C and A takes R hours.

Consider a route where we start at one of the airports, fly to another airport and then fly to the other airport.

What is the minimum possible sum of the flight times?

Constraints

1 \leq P,Q,R \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

P Q R

Output

Print the minimum possible sum of the flight times.

Sample Input 1

1 3 4

Sample Output 1

4

The sum of the flight times in the route A \rightarrow B \rightarrow C: 1 + 3 = 4 hours

The sum of the flight times in the route A \rightarrow C \rightarrow C: 4 + 3 = 7 hours

The sum of the flight times in the route B \rightarrow A \rightarrow C: 1 + 4 = 5 hours

The sum of the flight times in the route B \rightarrow C \rightarrow A: 3 + 4 = 7 hours

The sum of the flight times in the route C \rightarrow A \rightarrow B: 4 + 1 = 5 hours

The sum of the flight times in the route C \rightarrow B \rightarrow A: 3 + 1 = 4 hours

The minimum of these is 4 hours.

Sample Input 2

3 2 3

Sample Output 2

5

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

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;
 
static const double EPS = 1e-8;
static const double PI = 4.0 * atan(1.0);
static const ll INF = 1023456789;
 
#define REP(i,n)	for(int i=0;i<n;++i)
#define REPR(i,n)	for(int i=n-1;i>=0;--i)
#define FOR(i,s,n)	for(int i=s;i<n;++i)
#define FORR(i,s,n)	for(int i=n-1;i>=s;--i)
#define ALL(c)		(c).begin(),(c).end()
#define CLEAR(v)	memset(v,0,sizeof(v))
#define MP(a,b)		make_pair((a),(b))
#define ABS(a)		((a)>0?(a):-(a))
#define F			first
#define S			second
 
int main(int argc, char **argv) {
	int p, q, r;
	cin >> p >> q >> r;
	cout << min(p + q, min(q + r, r + p)) << endl;
	return 0;
}