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;
map<long long, long long> mp;
multiset<long long> ps;
multiset<pair<long long, long long> > usd;
long long f[4][int(3e5 + 500)];
int main() {
  long long n, k;
  cin >> n;
  for (int q = 0; q < 3; q++)
    for (int i = 0; i < 3 * n + 5; i++) f[q][i] = ((long long)(1e9 + 1));
  long long C = 1;
  for (int q = 0; q < 3; q++)
    for (int i = 0; i < n; i++) {
      cin >> k;
      if (!mp[k]) mp[k] = ++C;
      f[q][mp[k]] = min(f[q][mp[k]], i + 1ll);
    }
  long long pas = f[0][C];
  usd.insert({0, ((long long)(1e9 + 1)) + 5});
  usd.insert({((long long)(1e9 + 1)) + 5, 0});
  for (int i = C; i > 0; i--) {
    pas = min(pas, f[0][i] + *ps.begin());
    long long k, p;
    k = f[1][i];
    p = f[2][i];
    auto it = usd.lower_bound({k, p});
    if (it->second >= p) continue;
    ps.insert(it->second + k);
    long long lp = it->second;
    it--;
    while (it->second <= p) {
      ps.erase(ps.lower_bound(it->first + lp));
      lp = it->second;
      usd.erase(it--);
    }
    ps.erase(ps.lower_bound(it->first + lp));
    usd.insert({k, p});
    ps.insert(it->first + p);
  }
  cout << pas;
  return 0;
}
