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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
set<pair<int, int> > pos;
multiset<int> sum;
int a[3][100100], p[3][3 * 100100], arr[3 * 100100], id[3 * 100100];
bool cmp(int i, int j) { return p[0][i] > p[0][j]; }
void insert(pair<int, int> p) {
  set<pair<int, int> >::iterator l, m, r;
  r = pos.lower_bound(p);
  if (r->second >= p.second) return;
  l = m = pos.insert(p).first;
  l--;
  sum.erase(sum.find(l->first + r->second));
  sum.insert(m->first + r->second);
  while (l->second < m->second) {
    r = l;
    l--;
    sum.erase(sum.find(l->first + r->second));
    pos.erase(r);
  }
  sum.insert(l->first + m->second);
}
int main() {
  int n, m = 0;
  cin >> n;
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < n; j++) {
      scanf("%d", a[i] + j);
      arr[m++] = a[i][j];
    }
  }
  sort(arr, arr + m);
  m = unique(arr, arr + m) - arr;
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < n; j++)
      a[i][j] = lower_bound(arr, arr + m, a[i][j]) - arr;
  }
  memset(p, 0x3f, sizeof(p));
  for (int i = 0; i < 3; i++) {
    for (int j = n; j--;) p[i][a[i][j]] = j + 1;
  }
  for (int i = 0; i <= m; i++) id[i] = i;
  p[0][m] = 0;
  sort(id, id + m, cmp);
  pos.insert(pair<int, int>(0x3f3f3f3f + 1, 0));
  pos.insert(pair<int, int>(0, 0x3f3f3f3f + 1));
  sum.insert(0);
  int ans = p[0][id[0]];
  for (int i = 0; i < m; i++) {
    insert(pair<int, int>(p[1][id[i]], p[2][id[i]]));
    smin(ans, p[0][id[i + 1]] + *sum.begin());
  }
  cout << ans << endl;
  return 0;
}
