#include <bits/stdc++.h>
using namespace std;
const int MAXN=1000010;
int n,ans,flg,a[MAXN];
char c[MAXN];
int main () {
	scanf("%d%s",&n,c+1);
	for (int i=1;i<=n;i++) {
		a[i]=c[i]-'1';
		if (a[i]==1) {flg=1;}
	}
	if (!flg) {
		for (int i=0;i<=n-1;i++) {
			if (((n-1)&i)==i) {
				ans^=a[i+1];
			}
		}
		printf("%d",ans);
	} else {
		for (int i=1;i<=n;i++) {
			if (a[i]==2) {a[i]=0;}
		}
		for (int i=0;i<=n-1;i++) {
			if (((n-1)&i)==i) {
				ans^=a[i+1];
			}
		}
		printf("%d",ans);
	}
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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}.

Examples

Input

4
1231


Output

1


Input

10
2311312312


Output

0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN=1000010;
int n,ans,flg,a[MAXN];
char c[MAXN];
int main () {
	scanf("%d%s",&n,c+1);
	for (int i=1;i<=n;i++) {
		a[i]=c[i]-'1';
		if (a[i]==1) {flg=1;}
	}
	if (!flg) {
		for (int i=0;i<=n-1;i++) {
			if (((n-1)&i)==i) {
				ans^=a[i+1];
			}
		}
		printf("%d",ans);
	} else {
		for (int i=1;i<=n;i++) {
			if (a[i]==2) {a[i]=0;}
		}
		for (int i=0;i<=n-1;i++) {
			if (((n-1)&i)==i) {
				ans^=a[i+1];
			}
		}
		printf("%d",ans);
	}
	return 0;
}
```