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;
const int N = 1e5 + 50;
multiset<pair<int, int> > B;
multiset<int> A;
int n, m, a[N], b[N], c[N], ss[N * 3], fa[N], fb[N], fc[N], id[N], ans;
bool cmp(int x, int y) { return fa[x] > fa[y]; }
void ins(pair<int, int> nw) {
  set<pair<int, int> >::iterator i, j, k;
  k = B.lower_bound(nw);
  if ((*k).second >= nw.second) return;
  i = k;
  --i;
  A.erase(A.find((*i).first + (*k).second));
  A.insert(nw.first + (*k).second);
  while ((*i).second <= nw.second) {
    j = i;
    --i;
    A.erase(A.find((*i).first + (*j).second));
    B.erase(j);
  }
  A.insert((*i).first + nw.second);
  B.insert(nw);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), ss[++m] = a[i];
  for (int i = 1; i <= n; i++) scanf("%d", b + i), ss[++m] = b[i];
  for (int i = 1; i <= n; i++) scanf("%d", c + i), ss[++m] = c[i];
  sort(ss + 1, ss + m + 1);
  m = unique(ss + 1, ss + m + 1) - ss - 1;
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(ss + 1, ss + m + 1, a[i]) - ss;
    b[i] = lower_bound(ss + 1, ss + m + 1, b[i]) - ss;
    c[i] = lower_bound(ss + 1, ss + m + 1, c[i]) - ss;
    if (!fa[a[i]]) fa[a[i]] = i;
    if (!fb[b[i]]) fb[b[i]] = i;
    if (!fc[c[i]]) fc[c[i]] = i;
  }
  for (int i = 1; i <= m; i++) {
    if (!fa[i]) fa[i] = n * 3;
    if (!fb[i]) fb[i] = n * 3;
    if (!fc[i]) fc[i] = n * 3;
  }
  for (int i = 1; i <= m; i++) id[i] = i;
  sort(id + 1, id + m + 1, cmp);
  ans = fa[id[1]];
  fa[id[m + 1] = 0] = 0;
  A.insert(0);
  B.insert(make_pair(n * 10, 0));
  B.insert(make_pair(0, n * 10));
  for (int i = 1; i <= m; i++) {
    ins(make_pair(fb[id[i]], fc[id[i]]));
    ans = min(ans, fa[id[i + 1]] + (*A.begin()));
  }
  printf("%d\n", ans);
  return 0;
}
