#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 998244353;

int n;
char a[1000005];
int solve()
{
	int ret = 0;
	rep(i, n) if((n - 1 & i) == i) ret ^= a[i] & 1;
	return ret;
}

int main()
{
	scanf("%d%s", &n, a);
	rep(i, n) a[i] -= '1';
	
	bool ev = true;
	rep(i, n) if(a[i] == 1) ev = false;
	
	if(ev) {
		rep(i, n) a[i] /= 2;
		printf("%d\n", solve() * 2);
	} else printf("%d\n", solve());
	return 0;
}
### Prompt

Please provide a Cpp coded solution 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>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int MOD = 998244353;

int n;
char a[1000005];
int solve()
{
	int ret = 0;
	rep(i, n) if((n - 1 & i) == i) ret ^= a[i] & 1;
	return ret;
}

int main()
{
	scanf("%d%s", &n, a);
	rep(i, n) a[i] -= '1';
	
	bool ev = true;
	rep(i, n) if(a[i] == 1) ev = false;
	
	if(ev) {
		rep(i, n) a[i] /= 2;
		printf("%d\n", solve() * 2);
	} else printf("%d\n", solve());
	return 0;
}
```