Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, ans = 300005, a[300005], b[300005], c[300005], posa[300005],
       posb[300005], posc[300005];
set<pair<int, int> > s;
multiset<int> ss;
map<int, int> mp;
inline set<pair<int, int> >::iterator add(set<pair<int, int> >::iterator it) {
  return ++it;
}
inline set<pair<int, int> >::iterator sub(set<pair<int, int> >::iterator it) {
  return --it;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int main() {
  n = read() * 3;
  for (int x, i = 1; i <= n; ++i) {
    if (!mp[x = read()]) mp[x] = mp.size();
    a[i] = b[i] = c[i] = mp[x];
  }
  rotate(b + 1, b + n / 3 + 1, b + n + 1);
  rotate(c + 1, c + n * 2 / 3 + 1, c + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (!posa[a[i]]) posa[a[i]] = i;
    if (!posb[b[i]]) posb[b[i]] = i;
    if (!posc[c[i]]) posc[c[i]] = i;
  }
  s.insert(make_pair(0, n + 1));
  s.insert(make_pair(n + 1, 0));
  ss.insert(0);
  for (int i = n; i >= 1; --i) {
    if (i <= n) ans = min(ans, i + *ss.begin());
    if (i != posa[a[i]]) continue;
    pair<int, int> t = make_pair(posb[a[i]], posc[a[i]]);
    if (s.lower_bound(t)->second > t.second) continue;
    auto it = s.insert(t).first;
    ss.erase(ss.find(sub(it)->first + add(it)->second));
    for (auto j = sub(it); j->second < it->second; j = sub(s.erase(j)))
      ss.erase(ss.find(sub(j)->first + j->second));
    ss.insert(sub(it)->first + t.second);
    ss.insert(t.first + add(it)->second);
  }
  ans = min(ans, *ss.begin());
  cout << ans << endl;
  return 0;
}