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.
#include <bits/stdc++.h>
using namespace std;
set<pair<int, int> > tr;
multiset<int> an;
int a[300005], b[300005], c[300005], fira[300005], firb[300005], firc[300005],
    n, bin[300005], id[300005], m = 0;
int cmp(int x, int y) { return fira[x] > fira[y]; }
void ins(pair<int, int> u) {
  set<pair<int, int> >::iterator l, i, r = tr.lower_bound(u);
  if (r->second >= u.second) return;
  i = tr.insert(u).first;
  l = i;
  l--;
  an.erase(an.lower_bound(l->first + r->second));
  an.insert(i->first + r->second);
  while (l->second < i->second) {
    r = l--;
    an.erase(an.lower_bound(l->first + r->second));
    tr.erase(r);
  }
  an.insert(l->first + i->second);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), bin[m++] = a[i];
  for (int i = 0; i < n; i++) scanf("%d", &b[i]), bin[m++] = b[i];
  for (int i = 0; i < n; i++) scanf("%d", &c[i]), bin[m++] = c[i];
  sort(bin, bin + m);
  m = unique(bin, bin + m) - bin;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(bin, bin + m, a[i]) - bin;
    b[i] = lower_bound(bin, bin + m, b[i]) - bin;
    c[i] = lower_bound(bin, bin + m, c[i]) - bin;
    if (!fira[a[i]]) fira[a[i]] = i + 1;
    if (!firb[b[i]]) firb[b[i]] = i + 1;
    if (!firc[c[i]]) firc[c[i]] = i + 1;
  }
  for (int i = 0; i < m; i++) {
    if (!fira[i]) fira[i] = 5 * n;
    if (!firb[i]) firb[i] = 5 * n;
    if (!firc[i]) firc[i] = 5 * n;
  }
  for (int i = 0; i <= m; i++) id[i] = i;
  fira[m] = 0;
  sort(id, id + m, cmp);
  an.insert(0);
  int ans = fira[*id];
  tr.insert(make_pair(7 * n, 0));
  tr.insert(make_pair(0, 7 * n));
  for (int i = 0; i < m; i++) {
    ins(make_pair(firb[id[i]], firc[id[i]]));
    ans = min(ans, fira[id[i + 1]] + *an.begin());
  }
  cout << ans << endl;
  return 0;
}
