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.
1
2
3
2
 8
190409007 190409007 352375776 190409007 352375776 352375776 352375776 352375776
190409007 190409007 190409007 190409007 190409007 352375776 352375776 190409007
190409007 352375776 352375776 190409007 190409007 190409007 352375776 352375776
 1
1
1
1
 1
377489979
588153796
588153796
 2
1 1
2 2
3 3
 4
1 1 1 1
1 1 1 1
1 1 1 1
 2
1 2
2 2
1 1
 3
1 1 2
1 1 3
1 1 4

2
 2
 1
 2
 3
 1
 2
 9

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
const int INF = 2e8;
map<int, int> mp[4];
tuple<int, int, int> v[MAXN];
set<pair<int, int> > cl;
multiset<int> valores;
int n, respf;
void update(int a, int b) {
  auto mit = cl.lower_bound({a, b});
  if (mit->first >= a && mit->second >= b) return;
  auto bb = (*mit);
  auto aa = (*(--mit));
  valores.erase(valores.find(aa.first + bb.second));
  valores.insert(aa.first + b);
  valores.insert(a + bb.second);
  cl.insert({a, b});
  mit = cl.lower_bound({a, b});
  mit--;
  while (mit->first <= a && mit->second <= b) {
    bb = (*(mit));
    aa = (*(--mit));
    mit = cl.erase(++mit);
    mit--;
    valores.erase(valores.find(aa.first + bb.second));
    valores.erase(valores.find(bb.first + b));
    valores.insert(aa.first + b);
  }
}
int query() { return *valores.begin(); }
int main() {
  scanf("%d", &n);
  respf = 3 * n;
  for (int i = 1; i <= 3; i++) {
    for (int j = 1; j <= n; j++) {
      int val;
      scanf("%d", &val);
      if (mp[1][val] == 0) mp[1][val] = INF;
      if (mp[2][val] == 0) mp[2][val] = INF;
      if (mp[3][val] == 0) mp[3][val] = INF;
      mp[i][val] = min(mp[i][val], j);
    }
  }
  n = 0;
  for (auto &mit : mp[1]) {
    int val = mit.first;
    v[++n] = {mp[1][val], mp[2][val], mp[3][val]};
  }
  sort(v + 1, v + 1 + n);
  v[0] = {0, 0, 0};
  cl.insert({0, INF * 2});
  cl.insert({INF * 2, 0});
  valores.insert(0);
  respf = min(respf, get<0>(v[n]));
  for (int i = n; i >= 1; i--) {
    int custo = get<0>(v[i - 1]);
    update(get<1>(v[i]), get<2>(v[i]));
    custo += query();
    respf = min(respf, custo);
  }
  printf("%d\n", respf);
  return 0;
}
