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
#include <bits/stdc++.h>
using namespace std;
const int maxn = 22, maxm = (1 << 21) | 10;
const long long T = 93824992236885ll, S = (1ll << 50) - 1;
int m, n, PC[maxm];
char s[maxm];
long long a[maxm], b[maxm];
inline long long sum(long long x, long long y) {
  return x ^ y ^ ((x & y & T) << 1);
}
inline void upd(long long &x, const long long &y) {
  x ^= y ^ ((x & y & T) << 1);
  return;
}
inline void g(long long *a) {
  scanf("%s", s);
  for (int i = 0; i < n; ++i) a[i] = ((long long)s[i] - '0') << PC[i];
  for (int hl = 1, l = 2; l <= n; hl = l, l <<= 1)
    for (int i = 0; i < n; i += l)
      for (int j = 0; j < hl; ++j) upd(a[i | j | hl], a[i | j]);
  return;
}
inline void mul(long long &x, const long long &y) {
  long long Y[4];
  Y[0] = 0;
  Y[1] = y;
  Y[2] = sum(y, y);
  Y[3] = sum(Y[2], y);
  long long z = 0;
  int foo;
  while (x) {
    if (foo = (x & 3)) upd(z, Y[foo]);
    (Y[1] <<= 2) &= S;
    (Y[2] <<= 2) &= S;
    (Y[3] <<= 2) &= S;
    x >>= 2;
  }
  x = z;
  return;
}
int main() {
  scanf("%d", &m);
  n = 1 << m;
  for (int i = 0; i < n; ++i) PC[i] = __builtin_popcount(i) << 1;
  g(a);
  g(b);
  for (int i = 0; i < n; ++i) mul(a[i], b[i]);
  for (int hl = 1, l = 2; l <= n; hl = l, l <<= 1)
    for (int i = 0; i < n; i += l)
      for (int j = 0; j < hl; ++j) upd(a[i | j | hl], sum(T, S ^ a[i | j]));
  for (int i = 0; i < n; ++i) s[i] = ((a[i] >> PC[i]) & 3) + '0';
  fwrite(s, n, 1, stdout);
  return 0;
}
