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>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = mod * mod;
const int INF_N = 1e+9;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

//繰り返し二乗法
ll mod_pow(ll a, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%m;
		a = a * a%m; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//逆元(Eucledean algorithm)
ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}
using mP = pair<modint, modint>;

int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };




void solve() {
    vec a(5);
    rep(i, 5) cin >> a[i];
    int res = 0;
    rep(i, 5){
        if(a[i] == 0) res = i;
    }
    cout << res+1 << endl;
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  //cout << fixed << setprecision(10);
  //init_f();
  //init();
  //int t; cin >> t; rep(i, t)solve();
  solve();
//   stop
    return 0;
}