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 <bits/stdc++.h>
 
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
 
using namespace std;
 
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef double ld;
 
template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }



int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.precision(10);
	cout << fixed;
  
	long long a, b;
	cin >> a >> b;
	bitset<32> b1(a), b2(b);
	cout << (b1&b2) << '\n' << (b1|b2) << '\n' << (b1^b2) << '\n';
	//long long n;
	//cin >> n;
	//bitset<32> b(n);
	//cout << b << '\n' << (~b) << '\n' << (b<<1) << '\n' << (b>>1) << '\n';
    return 0;
}