#include <cstdio>
#include <algorithm>

const int MN = 1000005;

int N, A[MN], cnt[3], mul, Ans;

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i) scanf("%1d", &A[i]), --A[i];
//	for (int i = 1; i <= N; ++i) printf("%d", A[i]); puts("");
	for (int i = 1; i <= N; ++i) ++cnt[A[i]];
	if (cnt[0] == N) return puts("0"), 0;
	if (cnt[0] + cnt[2] == N) {
		for (int i = 1; i <= N; ++i) A[i] >>= 1;
		mul = 1;
	} else {
		for (int i = 1; i <= N; ++i) A[i] &= 1;
		mul = 0;
	}
//	for (int i = 1; i <= N; ++i) printf("%d", A[i]); puts("");
	for (int i = 1; i <= N; ++i) if (A[i] && !((i - 1) & (N - i))) Ans ^= 1;
	printf("%d\n", Ans << mul);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
#include <algorithm>

const int MN = 1000005;

int N, A[MN], cnt[3], mul, Ans;

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i) scanf("%1d", &A[i]), --A[i];
//	for (int i = 1; i <= N; ++i) printf("%d", A[i]); puts("");
	for (int i = 1; i <= N; ++i) ++cnt[A[i]];
	if (cnt[0] == N) return puts("0"), 0;
	if (cnt[0] + cnt[2] == N) {
		for (int i = 1; i <= N; ++i) A[i] >>= 1;
		mul = 1;
	} else {
		for (int i = 1; i <= N; ++i) A[i] &= 1;
		mul = 0;
	}
//	for (int i = 1; i <= N; ++i) printf("%d", A[i]); puts("");
	for (int i = 1; i <= N; ++i) if (A[i] && !((i - 1) & (N - i))) Ans ^= 1;
	printf("%d\n", Ans << mul);
	return 0;
}
```