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 = 100005;
const int INF = 1000000005;
int a[N], b[N], c[N];
set<int> tot;
set<pair<int, int> > dt;
map<int, int> A, B, C, rt;
void insert(int v, int w) {
  if (!v) v = INF;
  if (!w) w = INF;
  set<pair<int, int> >::iterator it, it1;
  it = dt.lower_bound(make_pair(v, 0));
  it--;
  if (it->second >= w) return;
  dt.insert(make_pair(v, it->second));
  rt[v + it->second]++;
  int pre, fg = 1;
  while (fg) {
    if (it->second >= w) break;
    pre = it->first;
    if (it == dt.begin()) fg = 0;
    pre = it->first;
    int s = it->first + it->second;
    it1 = it;
    if (fg) it--;
    dt.erase(*it1);
    rt[s]--;
    if (!rt[s]) rt.erase(s);
  }
  dt.insert(make_pair(pre, w));
  rt[pre + w]++;
}
int main() {
  int n, i;
  set<int>::iterator it;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    tot.insert(a[i]);
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    tot.insert(b[i]);
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    tot.insert(c[i]);
  }
  for (i = n; i > 0; i--) A[a[i]] = i, B[b[i]] = i, C[c[i]] = i;
  int ret = INF;
  dt.insert(make_pair(0, 0));
  rt[0] = 1;
  for (it = tot.begin(); it != tot.end(); it++)
    if (!A[*it]) {
      insert(B[*it], C[*it]);
    }
  for (i = n; i > 0; i--)
    if (A[a[i]] == i) {
      int tmp = i + rt.begin()->first;
      if (ret > tmp) ret = tmp;
      insert(B[a[i]], C[a[i]]);
    }
  int tmp = rt.begin()->first;
  if (ret > tmp) ret = tmp;
  printf("%d\n", ret);
  return 0;
}
