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;
multiset<int> li;
multiset<pair<int, int> > ha;
set<pair<int, int> >::iterator it, ti;
pair<int, int> a[100050 * 3];
int wz[100050 * 3][3], sg[100050 * 3], n, m, st, ans = 3 * 100050;
inline int Read() {
  int x = 0;
  char y;
  do y = getchar();
  while (y < '0' || y > '9');
  do x = x * 10 + y - '0', y = getchar();
  while (y >= '0' && y <= '9');
  return x;
}
void Pretreat() {
  sort(a + 1, a + n * 3 + 1);
  for (int i = 1; i <= n * 3; i++) {
    st++;
    int ri = i;
    while (ri < n * 3 && a[ri + 1].first == a[ri].first) ri++;
    for (int j = ri; j >= i; j--)
      if (a[j].second > n * 2)
        wz[st][2] = a[j].second - n * 2;
      else if (a[j].second > n)
        wz[st][1] = a[j].second - n;
      else
        wz[st][0] = a[j].second;
    i = ri;
  }
  for (int i = 1; i <= st; i++)
    for (int j = 0; j < 3; j++)
      if (!wz[i][j]) wz[i][j] = 100050 * 3;
  return;
}
inline bool cmp(int x, int y) { return wz[x][2] < wz[y][2]; }
inline void Erase(int x) { li.erase(li.lower_bound(x)); }
void Insert(int x) {
  it = ha.lower_bound(make_pair(wz[x][0], wz[x][1]));
  if (it->second >= wz[x][1]) return;
  ha.insert(make_pair(wz[x][0], wz[x][1]));
  it = ti = ha.lower_bound(make_pair(wz[x][0], wz[x][1]));
  pair<int, int> Now = *(++ti);
  while (true) {
    ti = it;
    ti--;
    if (ti->second <= it->second)
      Erase(ti->first + Now.second), Now = *ti, ha.erase(ti);
    else
      break;
  }
  Erase(ti->first + Now.second);
  ti = it;
  li.insert((--ti)->first + it->second);
  ti = it;
  li.insert((++ti)->second + it->first);
  return;
}
int Get() { return li.begin() == li.end() ? 0 : *li.begin(); }
void Solve() {
  li.insert(0);
  ha.insert(make_pair(0, 0x3f7f7f7f));
  ha.insert(make_pair(0x3f7f7f7f, 0));
  for (int i = 1; i <= st; i++) sg[i] = i;
  sort(sg + 1, sg + st + 1, cmp);
  for (int i = st; i >= 1; i--) {
    ans = min(ans, Get() + wz[sg[i]][2]);
    Insert(sg[i]);
    while (i - 1 && wz[sg[i]][2] == wz[sg[i - 1]][2]) Insert(sg[--i]);
  }
  ans = min(ans, Get());
}
int main() {
  n = Read();
  for (int j = 0; j < 3; j++)
    for (int i = 1; i <= n; i++) a[n * j + i] = make_pair(Read(), n * j + i);
  Pretreat();
  Solve();
  cout << ans << endl;
  return 0;
}
