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 <cassert>

using std::cin, std::cout, std::bitset, std::complex;
using std::vector, std::array, std::string, std::pair, std::list, std::queue;
using std::priority_queue, std::set, std::map, std::unordered_map;
using std::sort, std::min, std::max, std::reverse;
using std::min_element, std::max_element, std::accumulate;

using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using ull = unsigned long long int;
using ll = long long int;
using vll = vector<ll>; using vvll = vector<vll>; using vvvll = vector<vvll>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
using P = pair<int, int>;
using Pll = pair<ll, ll>;
using cdouble = complex<double>;

const double eps = 1e-7;
#define Loop(i, n) for(int i = 0; i < int(n); i++)
#define Loopll(i, n) for(ll i = 0; i < ll(n); i++)
#define Loop1(i, n) for(int i = 1; i <= int(n); i++)
#define Loopll1(i, n) for(ll i = 1; i <= ll(n); i++)
#define Loopr(i, n) for(int i = int(n) - 1; i >= 0; i--)
#define Looprll(i, n) for(ll i = ll(n) - 1; i >= 0; i--)
#define Loopr1(i, n) for(int i = int(n); i >= 1; i--)
#define Looprll1(i, n) for(ll i = ll(n); i >= 1; i--)
#define Foreach(buf, container) for(const auto &buf : container)
#define Foreachr(buf, container)  for(const auto &buf : reversed(container))
#define Loopdiag(i, j, h, w, sum) for(int i = ((sum) >= (h) ? (h) - 1 : (sum)), j = (sum) - i; i >= 0 && j < (w); i--, j++)
#define Loopdiagr(i, j, h, w, sum) for(int j = ((sum) >= (w) ? (w) - 1 : (sum)), i = (sum) - j; j >= 0 && i < (h); j--, i++)
#define Loopdiagsym(i, j, h, w, gap) for (int i = ((gap) >= 0 ? (gap) : 0), j = i - (gap); i < (h) && j < (w); i++, j++)
#define Loopdiagsymr(i, j, h, w, gap) for (int i = ((gap) > (h) - (w) - 1 ? (h) - 1 : (w) - 1 + (gap)), j = i - (gap); i >= 0 && j >= 0; i--, j--)
#define Loopitr(itr, container) for(auto itr = container.begin(); itr != container.end(); itr++)
#define quickio() std::ios::sync_with_stdio(false); std::cin.tie(0);
#define bitmanip(m,val) static_cast<bitset<(int)m>>(val)
#define Comp(type_t) bool operator<(const type_t &another) const
#define fst first
#define snd second
bool feq(double x, double y) { return abs(x - y) <= eps; }
bool inrange(ll x, ll t) { return x >= 0 && x < t; }
bool inrange(vll xs, ll t) { Foreach(x, xs) if (!(x >= 0 && x < t)) return false; return true; }
int ceillog2(ll x) { int m = int(log2(x)); return m + ((1LL << m) < x ? 1 : 0); }
int floorlog2(ll x) { int m = int(log2(x)); return m - ((1LL << m) > x ? 1 : 0); }
const string endl = "\n";
template<class T> T reversed(T container) { reverse(container.begin(), container.end()); return container; }
template<class T> void printv(const vector<T> &v) { for (const T &x : v) cout << x << " "; cout << endl; }
template<class T> void printmx(const vector<vector<T>> &mx) { for (const vector<T> &v : mx) printv(v); }
ll rndf(double x) { return (ll)(x + (x >= 0 ? 0.5 : -0.5)); }
ll floorsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (m*  m <= x ? 0 : -1); }
ll ceilsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (x <= m * m ? 0 : 1); }
ll rnddiv(ll a, ll b) { return (a / b + (a % b * 2 >= b ? 1 : 0)); }
ll ceildiv(ll a, ll b) { return (a / b + (a % b == 0 ? 0 : 1)); }
ll gcd(ll m, ll n) { if (n == 0) return m; else return gcd(n, m % n); }
ll lcm(ll m, ll n) { return ll(m) * ll(n) / gcd(m, n); }

//========================================================================//

int main() {
    int n = 5;
    vi x(n); Loop(i, n) cin >> x[i];
    Loop(i, n) {
        if (x[i] == 0) {
            cout << i + 1 << endl;
            return 0;
        }
    }
}