Naman has two binary strings s and t of length n (a binary string is a string which only consists of the characters "0" and "1"). He wants to convert s into t using the following operation as few times as possible.

In one operation, he can choose any subsequence of s and rotate it clockwise once.

For example, if s = 1110100, he can choose a subsequence corresponding to indices (1-based) \{2, 6, 7 \} and rotate them clockwise. The resulting string would then be s = 1010110.

A string a is said to be a subsequence of string b if a can be obtained from b by deleting some characters without changing the ordering of the remaining characters.

To perform a clockwise rotation on a sequence c of size k is to perform an operation which sets c_1:=c_k, c_2:=c_1, c_3:=c_2, …, c_k:=c_{k-1} simultaneously.

Determine the minimum number of operations Naman has to perform to convert s into t or say that it is impossible. 

Input

The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the strings.

The second line contains the binary string s of length n.

The third line contains the binary string t of length n.

Output

If it is impossible to convert s to t after any number of operations, print -1.

Otherwise, print the minimum number of operations required.

Examples

Input


6
010000
000001


Output


1

Input


10
1111100000
0000011111


Output


5

Input


8
10101010
01010101


Output


1

Input


10
1111100000
1111100001


Output


-1

Note

In the first test, Naman can choose the subsequence corresponding to indices \{2, 6\} and rotate it once to convert s into t.

In the second test, he can rotate the subsequence corresponding to all indices 5 times. It can be proved, that it is the minimum required number of operations.

In the last test, it is impossible to convert s into t.
#include <bits/stdc++.h>
using namespace std;
int n, i, j, k, a[1000005], b[1000005], c[2], cnt, p[1000005];
int t[1000005 * 4][2];
void maketree(int x, int l, int r) {
  if (l == r) {
    t[x][0] = t[x][1] = 0, t[x][p[l]] = l;
    return;
  }
  int mid = (l + r) >> 1;
  maketree(x << 1, l, mid), maketree(x << 1 ^ 1, mid + 1, r);
  t[x][0] = (t[x << 1][0]) ? t[x << 1][0] : t[x << 1 ^ 1][0];
  t[x][1] = (t[x << 1][1]) ? t[x << 1][1] : t[x << 1 ^ 1][1];
}
void change(int x, int l, int r, int v) {
  if (l == r) {
    t[x][0] = t[x][1] = 0;
    return;
  }
  int mid = (l + r) >> 1;
  if (v <= mid)
    change(x << 1, l, mid, v);
  else
    change(x << 1 ^ 1, mid + 1, r, v);
  t[x][0] = (t[x << 1][0]) ? t[x << 1][0] : t[x << 1 ^ 1][0];
  t[x][1] = (t[x << 1][1]) ? t[x << 1][1] : t[x << 1 ^ 1][1];
}
int find(int x, int l, int r, int L, int R, int v) {
  if (l > R || r < L) return 0;
  if (L <= l && r <= R) return t[x][v];
  int mid = (l + r) >> 1, tmp = find(x << 1, l, mid, L, R, v);
  if (tmp)
    return tmp;
  else
    return find(x << 1 ^ 1, mid + 1, r, L, R, v);
}
int main() {
  scanf("%d", &n);
  char ch = getchar();
  while (ch < '0' || ch > '1') ch = getchar();
  for (i = 1; i <= n; i++) a[i] = ch - '0', ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  for (i = 1; i <= n; i++) b[i] = ch - '0', ch = getchar();
  for (i = 1; i <= n; i++)
    if (a[i] != b[i]) p[++cnt] = a[i] > b[i];
  int res = 0, ans = 0;
  for (i = 1; i <= cnt; i++) res += p[i] ? 1 : -1;
  if (res != 0) printf("-1"), exit(0);
  if (!cnt) printf("0"), exit(0);
  maketree(1, 1, cnt);
  res = cnt;
  while (res) {
    ans++;
    int x = find(1, 1, cnt, 1, cnt, 0), y = find(1, 1, cnt, 1, cnt, 1), v,
        now = max(x, y);
    change(1, 1, cnt, x), change(1, 1, cnt, y), res -= 2;
    if (x > y)
      v = 1;
    else
      v = 0;
    if (!res) break;
    x = find(1, 1, cnt, now + 1, cnt, v);
    while (x) {
      y = find(1, 1, cnt, x + 1, cnt, v ^ 1);
      if (y) {
        change(1, 1, cnt, x), change(1, 1, cnt, y), res -= 2;
        now = y;
        if (!res) break;
        x = find(1, 1, cnt, now + 1, cnt, v);
      } else
        break;
    }
  }
  printf("%d\n", ans);
}
