#include<cstdio>
const int N=1000006;

int n,a[N],sum[N];
char s[N];
bool one=false;
bool odd=false;
bool two=false;

inline int Abs(int x) {
	return x<0?-x:x;
}
bool C2(int n,int m) {
	return sum[n]-sum[m]-sum[n-m];
}
int main() {
	scanf("%d",&n); n--;
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) {
		a[i]=Abs(s[i]-s[i+1]);
		if(!(i&1)) sum[i]=sum[i>>1]+1;
	}
	for(int i=1;i<=n;i++)
		sum[i]+=sum[i-1];
	for(int i=1;i<=n;i++) {
		if((a[i]&1)) {
			if(!C2(n-1,i-1))
				odd=!odd;
			one=true;
		} else if(a[i]==2)
			if(!C2(n-1,i-1))
				two=!two;
	}
	if(odd) puts("1");
	else if(one||!two) puts("0");
	else puts("2");
	return 0;
}
### Prompt

Create a solution in cpp for the following problem:

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<cstdio>
const int N=1000006;

int n,a[N],sum[N];
char s[N];
bool one=false;
bool odd=false;
bool two=false;

inline int Abs(int x) {
	return x<0?-x:x;
}
bool C2(int n,int m) {
	return sum[n]-sum[m]-sum[n-m];
}
int main() {
	scanf("%d",&n); n--;
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) {
		a[i]=Abs(s[i]-s[i+1]);
		if(!(i&1)) sum[i]=sum[i>>1]+1;
	}
	for(int i=1;i<=n;i++)
		sum[i]+=sum[i-1];
	for(int i=1;i<=n;i++) {
		if((a[i]&1)) {
			if(!C2(n-1,i-1))
				odd=!odd;
			one=true;
		} else if(a[i]==2)
			if(!C2(n-1,i-1))
				two=!two;
	}
	if(odd) puts("1");
	else if(one||!two) puts("0");
	else puts("2");
	return 0;
}
```