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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>   
using namespace std;
using namespace __gnu_pbds;
typedef long long ll; 
#define int long long
typedef unsigned long long lu;
typedef vector<ll> v;
typedef vector<vector<ll> > vv;
typedef vector<string>  vs;
typedef vector<pair<ll,ll>> vpr;
typedef long double ld;
#define f(i,n) for(ll i = 0; i < n; i++)
#define ff(i,n) for(ll i=1;i<=n;i++)
#define pb push_back
#define mp make_pair
#define endl "\n"
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define amax(x,y) if(y>x) x=y;
#define amin(x,y) if(y<x)x=y;
#define bg(x) x.begin()
#define sz(x) (ll)x.size()
#define in(x,n) for(ll i=0;i<n;i++)cin>>x[i]
#define out(x,n) for(ll i=0;i<n;i++)cout<<x[i]<<" " 
#define mxt(a) *(max_element(a.begin(),a.end()))
#define mnt(a) *(min_element(a.begin(),a.end())
#define tc ll t;cin>>t;while(t--)
typedef pair<ll,ll> pi;
#define yes cout<<"YES\n";
#define no cout<<"NO\n";
#define yesno(f) if(f) yes else no
const v dx = {1, -1, 0, 0};
const v dy = {0, 0, 1, -1};
const ld PI = 2 * acos(0.0);
ll cel(ll x1,ll y1){if((x1%y1)==0)return x1/y1;else return x1/y1+1;}
ll power(ll a,ll b,ll m)
{
    if(b==0)
        return 1;
    ll d=power(a,b/2,m);
    d=(d*d)%m;
    if(b&1)
        d=(d*a)%m;
    return d;
}
const ll mod=1e9+7;
int MOD(int a)
{
    if(a<0)
        a+=mod;
    if(a>=mod)
        a%=mod;
    return a;
}
// set_name.find_by_order(k) It returns to an iterator to the kth element (counting from zero) in the set in O(logn) time
// set_name.order_of_key(k) It returns to the number of items that are strictly smaller than our item k in O(logn) time.
/*string operations :
str.substr (x,y) : returns a substring str[x],str[x+1],...str[x+y-1]
str.substr (x) : returns a substring str[x],... end of string
str.find(qtr) : returns the first occurenece of qtr in str */
int32_t main()
{
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
    v a(5);in(a,5);
    f(i,5)
    {
        if(a[i]==0){
            cout<<i+1;return 0;
        }
    }
    return 0;
}