#include <bits/stdc++.h>
using namespace std;
void fwt(unsigned long long *p, int len) {
  for (int i = 1; i < len; i <<= 1)
    for (int j = 0; j < len; j++)
      if (j & i) p[j] += p[j ^ i];
}
void ifwt(unsigned long long *p, int len) {
  for (int i = 1; i < len; i <<= 1)
    for (int j = 0; j < len; j++)
      if (j & i) p[j] -= p[j ^ i];
}
const int Maxn = 1 << 21;
char s1[Maxn + 5], s2[Maxn + 5];
unsigned long long p[Maxn + 5], q[Maxn + 5];
int bit[Maxn + 5];
int main() {
  int n;
  scanf("%d%s%s", &n, s1, s2);
  for (int i = 0; i < (1 << n); i++) {
    bit[i] = bit[i >> 1] + ((i & 1) << 1);
    p[i] = ((unsigned long long)(s1[i] - '0') << bit[i]);
    q[i] = ((unsigned long long)(s2[i] - '0') << bit[i]);
  }
  fwt(p, 1 << n);
  fwt(q, 1 << n);
  for (int i = 0; i < (1 << n); i++) p[i] *= q[i];
  ifwt(p, 1 << n);
  for (int i = 0; i < (1 << n); i++) putchar(((p[i] >> bit[i]) & 3) + '0');
  putchar('\n');
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

Little C loves number «3» very much. He loves all things about it.

Now he is interested in the following problem:

There are two arrays of 2^n intergers a_0,a_1,...,a_{2^n-1} and b_0,b_1,...,b_{2^n-1}.

The task is for each i (0 ≤ i ≤ 2^n-1), to calculate c_i=∑ a_j ⋅ b_k (j|k=i and j\&k=0, where "|" denotes [bitwise or operation](https://en.wikipedia.org/wiki/Bitwise_operation#OR) and "\&" denotes [bitwise and operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND)).

It's amazing that it can be proved that there are exactly 3^n triples (i,j,k), such that j|k=i, j\&k=0 and 0 ≤ i,j,k ≤ 2^n-1. So Little C wants to solve this excellent problem (because it's well related to 3) excellently.

Help him calculate all c_i. Little C loves 3 very much, so he only want to know each c_i \& 3.

Input

The first line contains one integer n (0 ≤ n ≤ 21).

The second line contains 2^n integers in [0,3] without spaces — the i-th of them is a_{i-1}.

The third line contains 2^n integers in [0,3] without spaces — the i-th of them is b_{i-1}.

Output

Print one line contains 2^n integers in [0,3] without spaces — the i-th of them is c_{i-1}\&3. (It's obvious that c_{i}\&3 is in [0,3]).

Examples

Input

1
11
11


Output

12

Input

2
0123
3210


Output

0322

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void fwt(unsigned long long *p, int len) {
  for (int i = 1; i < len; i <<= 1)
    for (int j = 0; j < len; j++)
      if (j & i) p[j] += p[j ^ i];
}
void ifwt(unsigned long long *p, int len) {
  for (int i = 1; i < len; i <<= 1)
    for (int j = 0; j < len; j++)
      if (j & i) p[j] -= p[j ^ i];
}
const int Maxn = 1 << 21;
char s1[Maxn + 5], s2[Maxn + 5];
unsigned long long p[Maxn + 5], q[Maxn + 5];
int bit[Maxn + 5];
int main() {
  int n;
  scanf("%d%s%s", &n, s1, s2);
  for (int i = 0; i < (1 << n); i++) {
    bit[i] = bit[i >> 1] + ((i & 1) << 1);
    p[i] = ((unsigned long long)(s1[i] - '0') << bit[i]);
    q[i] = ((unsigned long long)(s2[i] - '0') << bit[i]);
  }
  fwt(p, 1 << n);
  fwt(q, 1 << n);
  for (int i = 0; i < (1 << n); i++) p[i] *= q[i];
  ifwt(p, 1 << n);
  for (int i = 0; i < (1 << n); i++) putchar(((p[i] >> bit[i]) & 3) + '0');
  putchar('\n');
  return 0;
}

```