#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
// #include <atcoder/all>
// using namespace atcoder;
using namespace std;

using ll = long long;
using pll = pair<ll, ll>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vpll = vector<pll>;
using vvpll = vector<vpll>;
#define rep(i, l, r) for (ll i = l; i < ll(r); i++)
#define rrep(i, l, r) for (ll i = ll(r-1); i >= l; i--)

#pragma region // doubling
template <typename T>
T fpow(T base, T id, ll exp, std::function<T(T, T)> bop);
template <typename T>
T fpow(T base, ll exp);

template <typename T>
T fpow(T base, ll exp) {
    T ans = 1;
    while (exp > 0) {
        if (exp & 1ll) {
            ans = ans * base;
        }
        base = base * base;
        exp >>= 1;
    }
    return ans;
}

template <typename T>
T fpow(T base, T id, ll exp, std::function<T(T, T)> bop) {
    T ans = id;
    while (exp > 0) {
        if (exp & 1ll) {
            ans = bop(ans, base);
        }
        base = bop(base, base);
        exp >>= 1;
    }
    return ans;
}
#pragma endregion


#pragma region
#pragma endregion



#include <atcoder/all>
using namespace atcoder;
using mint = modint998244353;

ll nanka(ll l, ll r) {
    return ((r - l) %3 + 3) % 3;
}

vll solve(vll a) {
    if (a.size() == 1) return a;
    else {
        vll next;
        rep(i, 0, a.size() - 2) {
            if (
                nanka(a[i], a[i+1]) == 1
                && nanka(a[i+1], a[i+2]) == 1
                && nanka(a[i+2], a[i]) == 1
            ) {
                next.push_back(1);
            }
            else 
            if (
                nanka(a[i], a[i+1]) == 2
                && nanka(a[i+1], a[i+2]) == 2
                && nanka(a[i+2], a[i]) == 2
            ) {
                next.push_back(2);
            }
            else {
                next.push_back(0);
            }
        }
        return solve(next);
    }
}

int main() {
    ll n; cin >> n;
    n = n * 2 + 1;
    vll a(n);
    rep(i, 0, n) {
        cin >> a[i];
    }
    if (n == 1) {
        cout << a[0] << endl;
    }
    else {
        if (a[0] == a[1] || a[1] == a[2] || a[2] == a[0]) {
            cout << 0 << endl;
        }
        else {
            bool ok = true;
            rep(i, 3, n) {
                if (i % 2 == 1 && a[i] != a[1]) {
                    ok = false;
                    break;
                }
                if (i % 4 == 0 && a[i] != a[0]) {
                    ok = false;
                    break;
                }
                if (i % 4 == 2 && a[i] != a[2]) {
                    ok = false;
                    break;
                }
            }
            if (!ok) {
                cout << 0 << endl;
            }
            else {
                a.resize((n % 4 == 1) ? 5 : 3);
                cout << solve(a)[0] << endl;
            }
        }
    }
}