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 NN = 303030;
vector<pair<int, int> > V[NN];
set<pair<int, int> > P;
map<int, int> id[3];
multiset<int> uni;
set<pair<int, int> >::iterator it, it1, it2;
int main() {
  int n, ans = 0x3f3f3f3f;
  scanf("%d", &n);
  for (int i = 0, j, k; i < 3; i++) {
    for (j = 1; j <= n; j++) {
      scanf("%d", &k);
      uni.insert(k);
      if (!id[i].count(k)) id[i][k] = j;
    }
  }
  pair<int, int> p;
  for (auto r : uni) {
    p.first = p.second = NN;
    if (id[1].count(r)) p.first = id[1][r];
    if (id[2].count(r)) p.second = id[2][r];
    if (id[0].count(r))
      V[id[0][r]].push_back(p);
    else
      V[NN - 1].push_back(p);
  }
  uni.clear();
  uni.insert(0);
  P.insert(pair<int, int>(NN + 1, 0));
  P.insert(pair<int, int>(0, NN + 1));
  for (int i = NN - 1; i >= 0; --i) {
    for (auto p : V[i]) {
      it = P.lower_bound(p);
      if (it->second >= p.second) continue;
      while (true) {
        it1 = it--;
        uni.erase(uni.find(it1->second + it->first));
        if (it->second > p.second) break;
      }
      it = --P.lower_bound(p);
      while (it->second <= p.second) {
        it1 = it--;
        P.erase(it1);
      }
      it = P.insert(p).first;
      it++;
      it2 = it;
      it--;
      it--;
      it1 = it;
      it++;
      uni.insert(it1->first + it->second);
      uni.insert(it->first + it2->second);
    }
    ans = min(ans, i + *uni.begin() - 1);
  }
  printf("%d\n", ans);
  return 0;
}
