For given three integers $a, b, c$, print the minimum value and the maximum value.

Constraints

* $-1,000,000,000 \leq a, b, c \leq 1,000,000,000$

Input

The input is given in the following format.


$a \; b \; c\;$


Three integers $a, b, c$ are given in a line.

Output

Print the minimum and maximum values separated by a space in a line.

Example

Input

4 5 3


Output

3 5
#include <bits/stdc++.h>

using ll=long long;
using namespace std;

#define rep_i(i, n) for (int i = 0; i < (n); ++i)
#define rep_ll(i, n) for (long long i = 0; i < (n); ++i)
#define r_rep_i(i, start, end) for (int i = (start); i < (end); ++i)
#define r_rep_ll(i, start, end) for (long long i = (start); i < (end); ++i)
#define debug_vi(v) copy((v).begin(), (v).end(), ostream_iterator<int>(cout, " "));
#define debug_vll(v) copy((v).begin(), (v).end(), ostream_iterator<long long>(cout, " "));
#define debug_vd(v) copy((v).begin(), (v).end(), ostream_iterator<double>(cout, " "));
#define sort_v(v) sort((v).begin(), (v).end()); //昇順
#define d_sort_vi(v) sort((v).begin(), (v).end(), greater<int>()); //降
#define d_sort_vll(v) sort((v).begin(), (v).end(), greater<long long>()); //降
#define d_sort_vd(v) sort((v).begin(), (v).end(), greater<double>()); //昇順
#define say(t) cout << (t) << endl;
#define sum_vi(v) accumulate((v).begin(), (v).end(), 0);
#define sum_vll(v)accumulate((v).begin(), (v).end(), 0LL);

ll gcd(ll a, ll b) { return b?gcd(b, a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a, b)*b;}

const int IINF=(1<<30);
const ll LLINF = 1LL<<60;

char BigAlphaToSmall(char a){
    if (a >= 65 && a <= 90){
        return a + 32;
    }
    return a;
}

char SmallAlphaToBig(char a){
    if (a >= 97 && a <= 122){
        return a - 32;
    }
    return a;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    vector<int> hoge(3);

    rep_i(i, 3) cin >> hoge[i];
    
    cout << *min_element(hoge.begin(), hoge.end()) << " " << *max_element(hoge.begin(), hoge.end()) << endl;
}
