There are three arrays a, b and c. Each of them consists of n integers. SmallY wants to find three integers u, v, w (0 ≤ u, v, w ≤ n) such that the following condition holds: each number that appears in the union of a, b and c, appears either in the first u elements of a, or in the first v elements of b, or in the first w elements of c. Of course, SmallY doesn't want to have huge numbers u, v and w, so she wants sum u + v + w to be as small as possible.

Please, help her to find the minimal possible sum of u + v + w.

Input

The first line contains a single integer n (1 ≤ n ≤ 105). The second line contains n space-separated integers a1, a2, ..., an — array a. The third line contains the description of array b in the same format. The fourth line contains the description of array c in the same format. The following constraint holds: 1 ≤ ai, bi, ci ≤ 109.

Output

Print a single integer — the minimum possible sum of u + v + w.

Examples

Input

3
1 1 101
1 2 1
3 2 1


Output

5

Input

5
1 1 2 2 3
2 2 4 3 3
3 3 1 1 1


Output

5

Note

In the first example you should choose u = 3, v = 0, w = 2. 

In the second example you should choose u = 1, v = 3, w = 1.
#include <bits/stdc++.h>
using namespace std;
int i, j, k, m, n, out;
map<int, int> a, b, c, p;
map<int, int>::iterator it, ip;
pair<int, int> pi;
multiset<int> ans;
multiset<int>::iterator it2;
int x[101000], y[101000], z[101000];
void read(map<int, int> &a, int x[]) {
  for (int i = 1; i <= n; i++) {
    scanf("%d", x + i);
    if (!a.count(x[i]))
      a[x[i]] = i;
    else
      x[i] = 0;
  }
}
void delans(int sum) {
  it2 = ans.find(sum);
  ans.erase(it2);
}
void addp(int x, int y) {
  map<int, int>::iterator it;
  it = p.lower_bound(x);
  pair<int, int> pii, pi = *it;
  if ((*it).second > y) return;
  ans.insert(x + (*it).second);
  for (it--; it != p.begin();) {
    pii = pi;
    pi = *it;
    it--;
    if (pi.second < y) {
      delans(pi.first + pii.second);
      p.erase(pi.first);
    } else
      break;
  }
  p[x] = y;
  delans(pi.first + pii.second);
  ans.insert(pi.first + y);
}
int main() {
  scanf("%d", &n);
  read(a, x);
  read(b, y);
  read(c, z);
  out = 3 * n;
  p.clear();
  p[-1] = 5001000;
  p[5001000] = 0;
  p[0] = 5001000;
  ans.insert(0);
  for (it = b.begin(); it != b.end(); it++) {
    pi = *it;
    if (a.count(pi.first)) continue;
    if (!c.count(pi.first))
      addp(pi.second, 3 * n);
    else
      addp(pi.second, c[pi.first]);
  }
  for (it = c.begin(); it != c.end(); it++) {
    pi = *it;
    if (a.count(pi.first) || b.count(pi.first)) continue;
    addp(3 * n, pi.second);
  }
  int u = n;
  while (!x[u]) u--;
  out = *(ans.begin()) + u;
  for (i = n; i >= 1; i--)
    if (x[i]) {
      u = i - 1;
      while (u > 0 && x[u] == 0) u--;
      int pb, pc;
      if (b.count(x[i]))
        pb = b[x[i]];
      else
        pb = 3 * n;
      if (c.count(x[i]))
        pc = c[x[i]];
      else
        pc = 3 * n;
      if (pb == 3 * n && pc == 3 * n) break;
      addp(pb, pc);
      out = min(out, *(ans.begin()) + u);
    }
  printf("%d\n", out);
  return 0;
}
