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 INF = 100000000;
struct Point {
  int x, y, z;
  Point() { x = y = z = INF; }
} P[100005];
bool cmp(Point a, Point b) { return a.x < b.x; }
map<int, Point> MP;
set<pair<int, int> > S;
multiset<int> costs;
void insert(pair<int, int> a) {
  S.insert(a);
  costs.insert(a.first + a.second);
}
void remove(set<pair<int, int> >::iterator it) {
  costs.erase(costs.find(it->first + it->second));
  S.erase(it);
}
int main() {
  int N, a, n = 0, ans = INF, minf = INF;
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &a);
    MP[a].x = min(MP[a].x, i);
  }
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &a);
    MP[a].y = min(MP[a].y, i);
  }
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &a);
    MP[a].z = min(MP[a].z, i);
  }
  for (auto i = MP.begin(); i != MP.end(); ++i) P[n++] = i->second;
  sort(P, P + n, cmp);
  insert(make_pair(0, 0));
  for (int i = n - 1; i >= 0; --i) {
    ans = min(ans, P[i].x + *costs.begin());
    auto it = S.lower_bound(make_pair(P[i].y, 0));
    --it;
    int maxv = it->second, start = -1;
    while (it->second <= P[i].z) {
      start = it->first;
      if (it == S.begin()) {
        remove(it);
        break;
      }
      remove(it--);
    }
    if (start != -1) {
      insert(make_pair(start, P[i].z));
      if (P[i].y != INF) {
        insert(make_pair(P[i].y, maxv));
      }
    }
  }
  ans = min(ans, *costs.begin());
  printf("%d\n", ans);
}