#include <bits/stdc++.h>

const int MAX_N = int(1e6) + 10;

int n;
char s[MAX_N];
int cnt[3];

int main() {
	scanf("%d%s", &n, s);
	bool have_one = false;
	for (int i = 0; i < n; ++i) {
		if (s[i] == '2')
			have_one = true;
		cnt[s[i] - '1'] += ((n - 1) & i) == i;
	}
	if (cnt[1] & 1)
		puts("1");
	else {
		if (!have_one && (cnt[2] & 1))
			puts("2");
		else
			puts("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 <bits/stdc++.h>

const int MAX_N = int(1e6) + 10;

int n;
char s[MAX_N];
int cnt[3];

int main() {
	scanf("%d%s", &n, s);
	bool have_one = false;
	for (int i = 0; i < n; ++i) {
		if (s[i] == '2')
			have_one = true;
		cnt[s[i] - '1'] += ((n - 1) & i) == i;
	}
	if (cnt[1] & 1)
		puts("1");
	else {
		if (!have_one && (cnt[2] & 1))
			puts("2");
		else
			puts("0");
	}
}
```