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;
const int MAXU = 300005;
const int MAXN = 100005;
struct triple {
  int v[3];
  bool operator<(const triple &y) const { return v[0] > y.v[0]; }
};
pair<int, int> sv[3][MAXN];
int N, M, av[MAXU];
triple fo[MAXU];
map<int, int> pf;
map<int, int> cd;
void erase(int v) {
  if (cd[v] > 1)
    cd[v]--;
  else
    cd.erase(v);
}
void insert(int v) { cd[v]++; }
void del(map<int, int>::iterator it) {
  erase(prev(it)->first + it->second);
  erase(it->first + next(it)->second);
  insert(prev(it)->first + next(it)->second);
  pf.erase(it);
}
void add(int x, int y) {
  auto it = pf.lower_bound(x);
  erase(prev(it)->first + it->second);
  insert(prev(it)->first + y);
  insert(x + it->second);
  pf[x] = y;
}
void ins(triple &t) {
  auto it = pf.lower_bound(t.v[1]);
  if (it->second < t.v[2]) {
    if (it->first == t.v[1]) {
      del(it);
    }
    add(t.v[1], t.v[2]);
    for (it = --pf.lower_bound(t.v[1]); it->second < t.v[2];) del(it--);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < N; j++) {
      cin >> sv[i][j].first;
      sv[i][j].second = j;
      av[M++] = sv[i][j].first;
    }
    sort(sv[i], sv[i] + N);
  }
  sort(av, av + M);
  M = unique(av, av + M) - av;
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < 3; j++) {
      auto it = lower_bound(sv[j], sv[j] + N, make_pair(av[i], -1));
      if (it == (sv[j] + N) || it->first != av[i])
        fo[i].v[j] = MAXU;
      else
        fo[i].v[j] = it->second + 1;
    }
  }
  sort(fo, fo + M);
  pf[0] = MAXU;
  pf[MAXU + 1] = 0;
  cd[0] = 1;
  int ans = MAXU;
  for (int i = 0; i < M; i++) {
    ans = min(ans, fo[i].v[0] + cd.begin()->first);
    ins(fo[i]);
  }
  cout << min(ans, cd.begin()->first) << endl;
  return 0;
}
