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 = 111111;
const int INF = 100000000;
int a[3][N];
vector<int> values;
int n, m;
int first[3][3 * N], visit[3 * N];
set<pair<int, int> > s;
map<int, int> del;
priority_queue<int> q;
void insert(int a, int b) {
  pair<int, int> tmp = make_pair(a, b);
  set<pair<int, int> >::iterator nxt = s.lower_bound(tmp);
  if (nxt->second >= b) return;
  q.push(-(nxt->second + a));
  s.insert(tmp);
  nxt = s.upper_bound(tmp);
  pair<int, int> last = *nxt;
  while (1) {
    set<pair<int, int> >::iterator it = s.find(tmp);
    --it;
    if (it->second <= b) {
      del[it->first + last.second]++;
      last = *it;
      s.erase(it);
    } else {
      del[it->first + last.second]++;
      q.push(-(it->first + b));
      break;
    }
  }
}
int getMin() {
  while (1) {
    int x = -q.top();
    if (del[x]) {
      del[x]--;
      q.pop();
      continue;
    }
    return x;
  }
}
void show() {
  for (set<pair<int, int> >::iterator it = s.begin(); it != s.end(); ++it) {
    printf("%d %d\n", it->first, it->second);
  }
  printf("end----------\n");
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 3; i++)
    for (int j = 1; j <= n; j++) {
      scanf("%d", &a[i][j]);
      values.push_back(a[i][j]);
    }
  sort(values.begin(), values.end());
  values.erase(unique(values.begin(), values.end()), values.end());
  m = (int)values.size();
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < m; j++) {
      first[i][j] = INF;
    }
    for (int j = n; j >= 1; j--) {
      a[i][j] =
          lower_bound(values.begin(), values.end(), a[i][j]) - values.begin();
      first[i][a[i][j]] = j;
    }
  }
  fill(visit, visit + m, 0);
  s.insert(make_pair(0, INF + 1));
  s.insert(make_pair(INF + 1, 0));
  q.push(0);
  for (int i = 1; i < 3; i++)
    for (int j = 1; j <= n; j++) {
      int key = a[i][j];
      if (first[0][key] == INF && !visit[key]) {
        insert(first[1][key], first[2][key]);
        visit[key] = 1;
      }
    }
  int ans = INF;
  for (int i = n; i >= 1; i--) {
    ans = min(ans, i + getMin());
    if (first[0][a[0][i]] == i) {
      if (first[1][a[0][i]] == INF && first[2][a[0][i]] == INF) break;
      insert(first[1][a[0][i]], first[2][a[0][i]]);
    }
    if (i == 1) {
      ans = min(ans, getMin());
    }
  }
  printf("%d\n", ans);
  return 0;
}