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;
using pii = pair<int, int>;
template <class U>
inline void smin(U& a, U b) {
  if (a > b) a = b;
}
const int NN = 3e5 + 10;
int n, m, ans;
int a[NN], b[NN], c[NN];
int A[NN], B[NN], C[NN];
int I[NN], T[NN];
multiset<int> uni;
set<pii> H;
void add(pii p) {
  set<pii>::iterator l, i, r = H.lower_bound(p);
  if (r->second >= p.second) return;
  i = H.insert(p).first;
  l = i;
  l--;
  uni.erase(uni.lower_bound(l->first + r->second));
  uni.insert(i->first + r->second);
  while (l->second < i->second) {
    r = l;
    l--;
    uni.erase(uni.lower_bound(l->first + r->second));
    H.erase(r);
  }
  uni.insert(l->first + i->second);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i), T[m++] = a[i];
  for (int i = 0; i < n; i++) scanf("%d", b + i), T[m++] = b[i];
  for (int i = 0; i < n; i++) scanf("%d", c + i), T[m++] = c[i];
  sort(T, T + m);
  m = unique(T, T + m) - T;
  for (int i = 0; i < n; i++) a[i] = lower_bound(T, T + m, a[i]) - T;
  for (int i = 0; i < n; i++) b[i] = lower_bound(T, T + m, b[i]) - T;
  for (int i = 0; i < n; i++) c[i] = lower_bound(T, T + m, c[i]) - T;
  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; i++) I[i] = i;
  A[m] = 0;
  sort(I, I + m, [&](int i, int j) { return A[i] > A[j]; });
  H.insert({0, 6 * n});
  H.insert({6 * n, 0});
  uni.insert(0);
  ans = A[I[0]];
  for (int i = 0; i < m; i++) {
    add({B[I[i]], C[I[i]]});
    smin(ans, A[I[i + 1]] + *uni.begin());
  }
  cout << ans << endl;
  return 0;
}
