Problem:
Score : 100 points

Problem Statement

We have five variables x_1, x_2, x_3, x_4, and x_5.

The variable x_i was initially assigned a value of i.

Snuke chose one of these variables and assigned it 0.

You are given the values of the five variables after this assignment.

Find out which variable Snuke assigned 0.

Constraints

The values of x_1, x_2, x_3, x_4, and x_5 given as input are a possible outcome of the assignment by Snuke.

Input

Input is given from Standard Input in the following format:

x_1 x_2 x_3 x_4 x_5

Output

If the variable Snuke assigned 0 was x_i, print the integer i.

Sample Input 1

0 2 3 4 5

Sample Output 1

1

In this case, Snuke assigned 0 to x_1, so we should print 1.

Sample Input 2

1 2 0 4 5

Sample Output 2

3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

//Common
using namespace std;using f64 = double;using i64=long long;using u64=unsigned long long;
template<typename T> using Arr=vector<T>;template<typename T> using Func=function<T>;template<typename T> using Str=basic_string<T>;
#define hfor(v, s, e) for(int v=(s);(s)<=v&&v<(e);++v)//half-opened range
#define hfori(v, s, e) for(int v=(e)-1;(s)<=v&&v<(e);--v)//inversion
#define hforo(v, s, e) int v=(s);for(;(s)<=v&&v<(e);++v)//out declaration
#define hforoi(v, s, e) int v=(e)-1; for(;(s)<=v&&v<(e);--v)
#define cfor(v, s, e) hfor(v,(s),(e)+1)//closed range
#define cfori(v, s, e) hfori(v,(s),(e)+1)
#define cforo(v, s, e) hforo(v,(s),(e)+1)
#define cforoi(v, s, e) hforoi(v,(s),(e)+1)
#define rep(v,x) hfor(v,0,(x))
#define repi(v,x) hfori(v,0,(x))
#define repo(v,x) hforo(v,0,(x))
#define all(x) (x).begin(),(x).end()
#define pushb push_back
#define pushf push_front
#define popb pop_back
#define popf pop_front
#define empl emplace
#define emplb emplace_back
#define emplf emplace_front
#define fi first
#define se second
#define cxp constexpr
#define PQ std::priority_queue

//Math
template<typename T> cxp T inf() { return numeric_limits<T>::max() / 2; }
template<typename T> inline T sq(T x){return x*x;}
cxp int lgc(int x){return 64-__builtin_clzll(x-1);}
cxp int lgf(int x){return 63-__builtin_clzll(x);}
int rd(int lb, int ub){static mt19937 rng(time(0)^i64(new int)); return uniform_int_distribution<int>(lb, ub-1)(rng);}
int rd(int ub=inf<int>()){return rd(0,ub);}
const f64 pi=acosl(-1), eps=1e-10;

//IO
template<typename T>ostream& operator<<(ostream& s, const Arr<T>& a) {for(auto i:a) cout<<i<<' '; return s;}
int getint(){int x;cin>>x;return x;}
char getchr(){char x;cin>>x;return x;}
#ifndef DEBUG
	//#pragma GCC optimize ("Ofast") //somtimes it become more slower
	auto __PRE_RUN__=(ios::sync_with_stdio(0), cin.tie(0), cout.tie(0),(cout<<fixed<<setprecision(11)), 0);
#endif

//Misc
int sz(const auto& x){return x.size();}
auto split(auto s, auto p){
	Arr<decltype(s)> ret;
	auto it1=s.begin();
	for(auto it2=it1; (it2 = find(it1,s.end(),p)) != s.end(); it1=it2+1)
		ret.pushb({it1, it2});
	ret.pushb({it1,s.end()});
	return ret;
}
template<typename T> T rev(const T& a){return {a.rbegin(),a.rend()};}
Arr<int> range(int n){ Arr<int> ret(n); rep(i,n)ret[i]=i; return ret; }
struct defer{ defer(auto f):f(f){} ~defer(){f();} function<void()> f; };
#define defer(x) auto _##__COUNTER__ = defer([&](){x;});
#define lam(expr, ...) [&](__VA_ARGS__){return expr;}
#define reduce accumulate

#define endl '\n'//not interactive?
#define int i64//MLE?
using pint = pair<int,int>;
using tint = tuple<int,int,int>;

signed main(){
	int a[5];
	rep(i,5)
		cin>>a[i];
	rep(i,5){
		if(!a[i])
			cout<<i+1<<endl;
	}
	return 0;
}