Problem: Given two non-negative decimal integers $a$ and $b$, calculate their AND (logical conjunction), OR (logical disjunction) and XOR (exclusive disjunction) and print them in binary representation of 32 bits.

Constraints

* $0 \leq a, b \leq 2^{32} - 1$

Input

The input is given in the following format.


$a \; b$


Output

Print results of AND, OR and XOR in a line respectively.

Example

Input

8 10


Output

00000000000000000000000000001000
00000000000000000000000000001010
00000000000000000000000000000010

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <numeric>
#include <climits>
#include <iterator>
#include <iomanip>
#include <stack>
#include <set>
#include <cstdio>
#include <bitset>
#include <functional>
#include <tuple>
#include <list>
#include <deque>
using namespace std;
typedef pair<int, int> P;

#define FOR(i, a, n) for (ll i = (ll)a; i<(ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)

typedef long long ll;
typedef vector<int> VI;


int main() {
    cin.tie(0);
   	ios::sync_with_stdio(false);
    
    ll a, b; cin >> a >> b;
    ll n = (a&b);
    vector<int> v;
    for(int i=0; i<32; ++i){
        v.push_back(n&1);
        n>>=1;
    }
    reverse(v.begin(), v.end());
    for(auto e: v) cout << e;
    cout << endl;
    
    v.clear();
    n = a|b;
    for(int i=0; i<32; ++i){
        v.push_back(n&1);
        n>>=1;
    }
    reverse(v.begin(), v.end());
    for(auto e: v) cout << e;
    cout << endl;
    
    v.clear();
    n = a^b;
    for(int i=0; i<32; ++i){
        v.push_back(n&1);
        n>>=1;
    }
    reverse(v.begin(), v.end());
    for(auto e: v) cout << e;
    cout << endl;
    
	return 0;
}