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: Time Limit Exceeded)

#include<bits/stdc++.h>

using namespace std;

int main(){
    int n;
    cin>>n;
    int seq;
    cin>>seq;
    int dp[n+1] = {0};
    for (int i=n-1; i>=1; i--){
        int temp = (seq/pow(10,(n)-i));
        dp[i] = temp%10;
    }
    dp[n] = seq%10;
    /*
    for(int i=1; i<=n;i++){
        cout<<dp[i];
    }
    cout<<endl;
    */
    for (int j=2; j<=n; j++){
        for (int i=1; i<=n-(j-1);i++){
            dp[i] = abs(dp[i]-dp[i+1]);
            //cout<<dp[i];
        }
        //cout<<endl;
    }
    cout<<dp[1];
    return 0;
}