1138_B. Circus_28539
Polycarp is a head of a circus troupe. There are n — an even number — artists in the troupe. It is known whether the i-th artist can perform as a clown (if yes, then c_i = 1, otherwise c_i = 0), and whether they can perform as an acrobat (if yes, then a_i = 1, otherwise a_i = 0).

Split the artists into two performances in such a way that: 

  * each artist plays in exactly one performance, 
  * the number of artists in the two performances is equal (i.e. equal to n/2), 
  * the number of artists that can perform as clowns in the first performance is the same as the number of artists that can perform as acrobats in the second performance. 

Input

The first line contains a single integer n (2 ≤ n ≤ 5 000, n is even) — the number of artists in the troupe.

The second line contains n digits c_1 c_2 … c_n, the i-th of which is equal to 1 if the i-th artist can perform as a clown, and 0 otherwise.

The third line contains n digits a_1 a_2 … a_n, the i-th of which is equal to 1, if the i-th artist can perform as an acrobat, and 0 otherwise.

Output

Print n/2 distinct integers — the indices of the artists that should play in the first performance.

If there are multiple answers, print any.

If there is no solution, print a single integer -1.

Examples

Input


4
0011
0101


Output


1 4


Input


6
000000
111111


Output


-1


Input


4
0011
1100


Output


4 3


Input


8
00100101
01111100


Output


1 2 3 6

Note

In the first example, one of the possible divisions into two performances is as follows: in the first performance artists 1 and 4 should take part. Then the number of artists in the first performance who can perform as clowns is equal to 1. And the number of artists in the second performance who can perform as acrobats is 1 as well.

In the second example, the division is not possible.

In the third example, one of the possible divisions is as follows: in the first performance artists 3 and 4 should take part. Then in the first performance there are 2 artists who can perform as clowns. And the number of artists in the second performance who can perform as acrobats is 2 as well.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
const int N = (2e5 + 20);
using namespace std;
int xd[5];
vector<int> x1;
vector<int> x2;
vector<int> x3;
vector<int> x4;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  string s1, s2;
  cin >> s1 >> s2;
  for (int i = 0; i < n; i++) {
    if (s1[i] == '0' and s2[i] == '0') {
      xd[0]++;
      x1.push_back(i + 1);
    }
    if (s1[i] == '0' and s2[i] == '1') {
      xd[1]++;
      x2.push_back(i + 1);
    }
    if (s1[i] == '1' and s2[i] == '0') {
      xd[2]++;
      x3.push_back(i + 1);
    }
    if (s1[i] == '1' and s2[i] == '1') {
      xd[3]++;
      x4.push_back(i + 1);
    }
  }
  int a = -1, b, c, d;
  for (int i = 0; i <= xd[1]; i++) {
    for (int j = 0; j <= xd[2]; j++) {
      int u = xd[1] + xd[3] - i - j;
      int v = n / 2 - i - j - u / 2;
      if (u % 2 == 0 and u / 2 >= 0 and u / 2 <= xd[3] and v >= 0 and
          v <= xd[0]) {
        b = i;
        c = j;
        d = u / 2;
        a = v;
        break;
      }
    }
  }
  if (a == -1) {
    cout << -1 << endl;
    return 0;
  } else {
    for (int i = 0; i < a; i++) cout << x1[i] << " ";
    for (int i = 0; i < b; i++) cout << x2[i] << " ";
    for (int i = 0; i < c; i++) cout << x3[i] << " ";
    for (int i = 0; i < d; i++) cout << x4[i] << " ";
    cout << endl;
  }
  return 0;
}

4
0011
1100
 4
0011
0101
 8
00100101
01111100
 6
000000
111111
 10
0011101001
1000101110
 4
0011
1100
 6
111100
111000
 8
11111111
11110000
 2
10
10
 8
11100000
00101111
 2
10
01
 8
01111111
10000111
 10
1111000011
1111111100
 2
01
11
 10
0011101001
1010101110
 4
0001
1100
 6
111110
111000
 8
11111111
11110010
 8
11100000
00111111
 8
11111111
10000111
 10
1011000011
1111111100
 4
0011
1101
 8
00100101
01111110
 10
0011101001
1010111110
 6
111110
111010
 8
11011111
11110010
 8
11101000
00111111
 8
11011111
10000111
 4
0011
1001
 10
0011101001
1010111111
 8
11011101
11110010
 8
11101000
00101111
 8
10011111
10000111
 4
0011
1010
 10
0011101001
1010101111
 8
11011101
11111010
 8
11001000
00101111
 8
00011111
10000111
 4
0011
1110
 6
100101
111110
 10
0011101001
1010101011
 6
110110
011000
 10
1001001010
1111111100
 8
00010111
11111110
 6
110101
111110
 10
0011101001
1010101010
 8
11010101
11011010
 8
00010111
00000111
 10
1001011010
1111111100
 6
110101
011110
 8
11010101
11011011
 8
00010111
00100111
 10
1001011010
1111011100
 6
110101
001110
 4
0011
1000
 6
000100
111111
 4
0001
1101
 10
1011000010
1111111100
 4
0010
1101
 8
00100111
01111110
 6
000101
111111
 4
0001
1111
 6
111110
011010
 10
1011000010
1111111101
 4
0001
0101
 8
00110111
01111110
 6
000101
111110
 4
0000
1111
 6
111110
011000
 10
1011001010
1111111100
 4
0001
1110
 8
00010111
01111110
 4
0000
1110
 8
11011101
11011010
 8
11001000
00001111
 8
00010111
10000111
 4
0011
0110
 4
1001
1110
 4
0100
1110
 6
111110
001000
 8
11001000
00001110
 4
0010
0110
 4
1000
1110
 8
00010111
11110110
 10
0011101101
1010101010
 4
1100
1110
 6
011110
001000
 8
11001000
00101110
 4
0011
0100
 4
1000
1010
 8
00010101
11110110
 10
0011101101
1010101110
 4
1101
1110
 6
011110
001100
 8
11010101
01011011

1 2  1 4  2 3 4 5
 -1 1 2 5 8 9
 1 2  1 4 5  5 6 7 8  -1 3 5 6 7
 2  2 3 4 5  1 2 3 5 6
 2  1 2 3 5 8  1 2  1 4 6  1 5 6 8  -1
 2 3 4 5  1 2 3 4 5  1 4  2 3 4 6  1 3 5 6 8  1 2 6  1 3 5 6  3 4 5 6  1 2 3 4  2 4  1 3 5 6 7  1 3 5 7  3 6 7 8  1 2 3 6  2 3  1 3 5 8 9  1 2 3 7  3 5 6 7  2 3 6 7  1 3  1 2 4  1 2 3 5 9  1 3 6  1 2 3 4 7  1 4 6 7  1 2 3  1 2 3 5 6  1 2 3 5  1 2 4 6  1 2 3 4 6  2 3 5  1 2 5 7  1 2 6 7  1 2 3 4 8  1 3 5  1 2  -1
 1 4  1 2 3 4 5  -1
 2 3 4 6  -1
 -1
 1 2 6  -1
 1 4  2 3 4 5  -1
 -1
 1 4 6  1 2 3 4 5  -1
 2 3 4 6  -1
 1 2 3 7  3 5 6 7  2 3 6 7  1 3  1 2  1 2  -1
 3 4 5 6  1 3  1 2  1 2 4 6  1 2 3 5 6  1 3  1 2 6  3 4 5 6  1 2  1 2  1 2 4 6  1 2 3 5 9  1 3  1 3 6  2 3 4 5 