Regular Bracket Sequence

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

A string is called bracket sequence if it does not contain any characters other than “(” and “)”. A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters “+” and “1” into this sequence. For example, “”, “(())” and “()()” are regular bracket sequences; “))” and “)((” are bracket sequences (but not regular ones), and “(a)” and “(1)+(1)” are not bracket sequences at all.

You have a number of strings; each string is a bracket sequence of length 222. So, overall you have cnt1cnt1cnt1 strings “((”, cnt2cnt2cnt2 strings “()”, cnt3cnt3cnt3 strings “)(” and cnt4cnt4cnt4 strings “))”. You want to write all these strings in some order, one after another; after that, you will get a long bracket sequence of length 2(cnt1+cnt2+cnt3+cnt4)2(cnt1+cnt2+cnt3+cnt4)2(cnt1+cnt2+cnt3+cnt4). You wonder: is it possible to choose some order of the strings you have such that you will get a regular bracket sequence? Note that you may not remove any characters or strings, and you may not add anything either.

The input consists of four lines, iii-th of them contains one integer cnticnt_icnti​ (0≤cnti≤109)(0≤cnt_i≤10^9)(0≤cnti​≤109).

Print one integer: 111 if it is possible to form a regular bracket sequence by choosing the correct order of the given strings, 000 otherwise.

input

3
1
4
3

output

1

input

0
0
0
0

output

1

input

1
2
3
4

output

0

In the first example it is possible to construct a string “(())()(()((()()()())))”, which is a regular bracket sequence.

In the second example it is possible to construct a string “”, which is a regular bracket sequence.

你有四种括号，"(("，"()"，")("，"))"， 给出四种括号的数量， 问能否将这些括号以某种顺序连接起来， 使得每个左括号都有与之匹配的右括号。

观察可以发现：如果要完全匹配，()是不会对结果产生影响的，然后可以推出一个关系式：cnt1=2×cnt3+cnt4,cnt4=2×cnt3+cnt1cnt1=2\times cnt3+cnt4,cnt4=2\times cnt3+cnt1cnt1=2×cnt3+cnt4,cnt4=2×cnt3+cnt1

即：当cnt3≠0,cnt1=cnt4≠0cnt3\neq0,cnt1=cnt4\neq0cnt3̸​=0,cnt1=cnt4̸​=0或cnt3=0,cnt1=cnt2cnt3=0,cnt1=cnt2cnt3=0,cnt1=cnt2时，才会完全匹配

/*************************************************************************

     > Author: WZY
     > School: HPU
     > Created Time:   2019-03-15 10:34:53

************************************************************************/
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstring>
#include <limits.h>
#include <iostream>
#include <algorithm>
#include <random>
#include <iomanip>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <random>
#define ll long long
#define ull unsigned long long
#define lson o<<1
#define rson o<<1|1
#define ms(a,b) memset(a,b,sizeof(a))
#define SE(N) setprecision(N)
#define PSE(N) fixed<<setprecision(N)
#define bug cerr<<"-------------"<<endl
#define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"\n"
#define LEN(A) strlen(A)
const double E=exp(1);
const double eps=1e-9;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int maxn=1e6+10;
const int maxm=1e3+10;
const int moha=19260817;
const int inf=1<<30;
const ll INF=1LL<<60;
using namespace std;
inline void Debug(){cerr<<'\n';}
inline void MIN(int &x,int y) {if(y<x) x=y;}
inline void MAX(int &x,int y) {if(y>x) x=y;}
inline void MIN(ll &x,ll y) {if(y<x) x=y;}
inline void MAX(ll &x,ll y) {if(y>x) x=y;}
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
    cerr<<arg<<"";Debug(rest...);}
int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(false);cin.tie(0);
    cout.precision(20);
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    int cnt1,cnt2,cnt3,cnt4;
    while(cin>>cnt1>>cnt2>>cnt3>>cnt4)
    {
        if(cnt1+cnt4==0)
        {
            if(cnt3)
                cout<<0<<endl;
            else
                cout<<1<<endl;
            continue;
        }
        if(cnt1==cnt4)
            cout<<1<<endl;
        else
            cout<<0<<endl;
    }
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s.\n";
    #endif
    return 0;
}