Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n, m, ans, a[N], b[N], c[N], id[N], V[N], A[N], B[N], C[N];
multiset<int> res;
set<pair<int, int> > H;
bool cmp(const int i, const int j) { return A[i] > A[j]; }
void ins(pair<int, int> z) {
  set<pair<int, int> >::iterator l, i, r = H.lower_bound(z);
  if (r->second >= z.second) return;
  i = H.insert(z).first, l = i, --l,
  res.erase(res.lower_bound(l->first + r->second)),
  res.insert(i->first + r->second);
  while (l->second < i->second)
    r = l--, res.erase(res.lower_bound(l->first + r->second)), H.erase(r);
  res.insert(l->first + i->second);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i), V[m++] = a[i];
  for (int i = 0; i < n; ++i) scanf("%d", b + i), V[m++] = b[i];
  for (int i = 0; i < n; ++i) scanf("%d", c + i), V[m++] = c[i];
  sort(V, V + m), m = unique(V, V + m) - V;
  for (int i = 0; i < n; ++i) a[i] = lower_bound(V, V + m, a[i]) - V;
  for (int i = 0; i < n; ++i) b[i] = lower_bound(V, V + m, b[i]) - V;
  for (int i = 0; i < n; ++i) c[i] = lower_bound(V, V + m, c[i]) - V;
  for (int i = 0; i < m; ++i) A[i] = B[i] = C[i] = 5 * n;
  for (int i = 0; i < n; ++i)
    if (A[a[i]] > n) A[a[i]] = i + 1;
  for (int i = 0; i < n; ++i)
    if (B[b[i]] > n) B[b[i]] = i + 1;
  for (int i = 0; i < n; ++i)
    if (C[c[i]] > n) C[c[i]] = i + 1;
  for (int i = 0; i < m + 1; ++i) id[i] = i;
  A[m] = 0, sort(id, id + m, cmp);
  H.insert(make_pair(0, 6 * n)), H.insert(make_pair(6 * n, 0)), res.insert(0),
      ans = A[*id];
  for (int i = 0; i < m; ++i)
    ins(make_pair(B[id[i]], C[id[i]])),
        ans = min(ans, A[id[i + 1]] + *res.begin());
  printf("%d\n", ans);
  return 0;
}