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 MAX = 100005;
const int MOD = 1000000007;
const int MAGIC = 3;
struct trip {
  int a, b, c;
  trip() { a = b = c = 3 * MAX + 1; }
  bool operator<(trip t) const { return a > t.a; }
};
map<int, trip> moo;
vector<trip> data;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int k;
    scanf("%d", &k);
    moo[k].a = min(moo[k].a, i + 1);
  }
  for (int i = 0; i < n; i++) {
    int k;
    scanf("%d", &k);
    moo[k].b = min(moo[k].b, i + 1);
  }
  for (int i = 0; i < n; i++) {
    int k;
    scanf("%d", &k);
    moo[k].c = min(moo[k].c, i + 1);
  }
  for (pair<int, trip> i : moo) {
    data.push_back(i.second);
  }
  sort(data.begin(), data.end());
  set<pair<int, int> > second;
  second.insert({0, 0});
  multiset<int> sums;
  sums.insert(0);
  int mini = data[0].a;
  for (int j = 0; j < data.size(); j++) {
    int val = 0;
    if (j != data.size() - 1) {
      val = data[j + 1].a;
    }
    trip i = data[j];
    auto it = second.lower_bound({i.b, MOD});
    it--;
    auto up = it;
    if (up->second >= i.c) {
      mini = min(mini, val + *sums.begin());
      continue;
    }
    while (1) {
      if (it->second > i.c) {
        it++;
        break;
      }
      if (it == second.begin()) {
        break;
      }
      it--;
    }
    auto dn = it;
    pair<int, int> new1 = {i.b, up->second};
    pair<int, int> new2 = {dn->first, i.c};
    up++;
    while (dn != up) {
      sums.erase(sums.find(dn->first + dn->second));
      second.erase(dn++);
    }
    second.insert(new1);
    second.insert(new2);
    sums.insert(new1.first + new1.second);
    sums.insert(new2.first + new2.second);
    mini = min(mini, val + *sums.begin());
  }
  cout << mini << endl;
}