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;
int i, j, m, n, p, k, ans, a[400005], b[400005], c[400005], P[400005],
    A[400005], B[400005], C[400005], id[400005];
set<pair<int, int> > st;
multiset<int> Ans;
inline bool cmp(int a, int b) { return A[a] > A[b]; }
void ins(pair<int, int> now) {
  set<pair<int, int> >::iterator i, j, k;
  k = st.upper_bound(now);
  if (k->second >= now.second) return;
  i = st.insert(now).first;
  j = i;
  --j;
  Ans.erase(Ans.lower_bound(j->first + k->second)),
      Ans.insert(i->first + k->second);
  while (j->second < i->second)
    k = j--, Ans.erase(Ans.lower_bound(j->first + k->second)), st.erase(k);
  Ans.insert(j->first + i->second);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), P[++P[0]] = a[i];
  for (i = 1; i <= n; ++i) scanf("%d", &b[i]), P[++P[0]] = b[i];
  for (i = 1; i <= n; ++i) scanf("%d", &c[i]), P[++P[0]] = c[i];
  sort(P + 1, P + P[0] + 1);
  P[0] = unique(P + 1, P + P[0] + 1) - (P + 1);
  for (i = 1; i <= n; ++i) a[i] = lower_bound(P + 1, P + P[0] + 1, a[i]) - P;
  for (i = 1; i <= n; ++i) b[i] = lower_bound(P + 1, P + P[0] + 1, b[i]) - P;
  for (i = 1; i <= n; ++i) c[i] = lower_bound(P + 1, P + P[0] + 1, c[i]) - P;
  for (i = 1; i <= P[0]; ++i) id[i] = i, A[i] = B[i] = C[i] = 5 * n;
  for (i = 1; i <= n; ++i)
    if (A[a[i]] > n) A[a[i]] = i;
  for (i = 1; i <= n; ++i)
    if (B[b[i]] > n) B[b[i]] = i;
  for (i = 1; i <= n; ++i)
    if (C[c[i]] > n) C[c[i]] = i;
  st.insert(make_pair(0, 6 * n));
  st.insert(make_pair(6 * n, 0));
  id[P[0] + 1] = P[0] + 1;
  sort(id + 1, id + P[0] + 1, cmp);
  Ans.insert(0);
  ans = A[id[1]];
  for (i = 1; i <= P[0]; ++i) {
    ins(make_pair(B[id[i]], C[id[i]]));
    ans = min(ans, A[id[i + 1]] + *Ans.begin());
  }
  printf("%d\n", ans);
}
