Problem:
Score : 700 points

Problem Statement

Given is a sequence of N digits a_1a_2\ldots a_N, where each element is 1, 2, or 3.
Let x_{i,j} defined as follows:

x_{1,j} := a_j \quad (1 \leq j \leq N)

x_{i,j} := | x_{i-1,j} - x_{i-1,j+1} | \quad (2 \leq i \leq N and 1 \leq j \leq N+1-i)

Find x_{N,1}.

Constraints

2 \leq N \leq 10^6

a_i = 1,2,3 (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N
a_1a_2\ldotsa_N

Output

Print x_{N,1}.

Sample Input 1

4
1231

Sample Output 1

1

x_{1,1},x_{1,2},x_{1,3},x_{1,4} are respectively 1,2,3,1.

x_{2,1},x_{2,2},x_{2,3} are respectively |1-2| = 1,|2-3| = 1,|3-1| = 2.

x_{3,1},x_{3,2} are respectively |1-1| = 0,|1-2| = 1.

Finally, x_{4,1} = |0-1| = 1, so the answer is 1.

Sample Input 2

10
2311312312

Sample Output 2

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

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                       (| -_- |)
                       O\  =  /O
                    ____/`---'\____
                  .'  \\|     |//  `.
                 /  \\|||  :  |||//  \
                /  _||||| -:- |||||_  \
                |   | \\\  -  /'| |   |
                | \_|  `\`---'//  |_/ |
                \  .-\__ `-. -'__/-.  /
              ___`. .'  /--.--\  `. .'___
           ."" '<  `.___\_<|>_/___.' _> \"".
          | | :  `- \`. ;`. _/; .'/ /  .' ; |
          \  \ `-.   \_\_`. _.'_/_/  -' _.' /
===========`-.`___`-.__\ \___  /__.-'_.'_.-'================

                  Please give me AC.
*/

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <numeric>
#include <string>
#include <sstream>
#include <complex>
#include <bitset>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <chrono>
#include <random>

using namespace std;

using int64 = long long;
using uint64 = unsigned long long;
using vi = vector<int>;
using vl = vector<int64>;
using pii = pair<int, int>;
using pll = pair<int64, int64>;

#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define all(v) (v).begin(), (v).end()
#define print(x) cout << (x) << '\n'
#define print2(x, y) cout << (x) << ' ' << (y) << '\n'
#define print3(x, y, z) cout << (x) << ' ' << (y) << ' ' << (z) << '\n'
#define printn(v) rep(i, (v).size() - 1) cout << (v)[i] << ' '; cout << (v)[n - 1] << '\n';
#define debug(x) cerr << #x << ": " << (x) << '\n'
#define debug2(x, y) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << '\n'
#define debug3(x, y, z) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << ", " << #z << ": " << (z) << '\n'
#define dbg(v) for (size_t _ = 0; _ < v.size(); ++_){cerr << #v << "[" << _ << "] : " << v[_] << '\n';}
// constant
const int INF = (1<<30) - 1;
const int64 INF64 = (1LL<<62) - 1;

template<typename T> T gcd(T a, T b) {
    if (a < b) return gcd(b, a);
    T r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

template<typename T> T lcm(const T a, const T b) {
    return a / gcd(a, b) * b;
}

template<typename T> bool chmin(T& a, const T& b) {
    if (a > b) return a = b, true; else return false;
}

template<typename T> bool chmax(T& a, const T& b) {
    if (a < b) return a = b, true; else return false;
}

// End of template.
// begin fast io

char buf[1010000];
int bufi = 0, outi = 0;

template<typename T>
void readui(T &n) {
    n = 0;
    int c = buf[bufi++];
    for (; '0' <= c && c <= '9'; c = buf[bufi++])
        n = 10 * n + c - '0';
}

// end fast io

char Cmod2(int n, int r) {
    return n == (r | (n - r)) ? 3 : 0;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cin.read(buf, sizeof buf); // 注意: ./a.out < in か pbp | ./a.out で入力すること

    int n;
    readui(n);
    char *a = buf + bufi;
    rep(i, n) a[i] -= '1';

    char x = 0;
    rep(i, n) x ^= Cmod2(n - 1, i) & a[i];
    if (x & 1) {
        print(1);
        return 0;
    }

    rep(i, n) if (a[i] == 1) {
        print(0);
        return 0;
    }

    if (x & 2) {
        print(2);
    } else {
        print(0);
    }
    return 0;
}