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.
1
2
3
2
 8
190409007 190409007 352375776 190409007 352375776 352375776 352375776 352375776
190409007 190409007 190409007 190409007 190409007 352375776 352375776 190409007
190409007 352375776 352375776 190409007 190409007 190409007 352375776 352375776
 1
1
1
1
 1
377489979
588153796
588153796
 2
1 1
2 2
3 3
 4
1 1 1 1
1 1 1 1
1 1 1 1
 2
1 2
2 2
1 1
 3
1 1 2
1 1 3
1 1 4

2
 2
 1
 2
 3
 1
 2
 9

#include <bits/stdc++.h>
using namespace std;
struct data {
  int a, b, c;
  bool operator<(const data &rhs) const {
    return min(a, min(b, c)) > min(rhs.a, min(rhs.b, rhs.c));
  }
} D[300005];
vector<int> v;
int n, nn;
int a[100005];
int b[100005];
int c[100005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), v.push_back(a[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", b + i), v.push_back(b[i]);
  for (int i = 1; i <= n; ++i) scanf("%d", c + i), v.push_back(c[i]);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  nn = v.size();
  for (int i = 0; i < nn; ++i) {
    D[i].a = D[i].b = D[i].c = n + 1;
  }
  for (int i = 1; i <= n; ++i) {
    int a_idx = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
    int b_idx = lower_bound(v.begin(), v.end(), b[i]) - v.begin();
    int c_idx = lower_bound(v.begin(), v.end(), c[i]) - v.begin();
    D[a_idx].a = min(D[a_idx].a, i);
    D[b_idx].b = min(D[b_idx].b, i);
    D[c_idx].c = min(D[c_idx].c, i);
  }
  sort(D, D + nn);
  {
    int u = 0, v = 0, w = 0;
    for (int i = 0; i < nn; ++i) {
      if (D[i].a == n + 1 && D[i].b == n + 1) w = max(w, D[i].c);
      if (D[i].b == n + 1 && D[i].c == n + 1) u = max(u, D[i].a);
      if (D[i].c == n + 1 && D[i].a == n + 1) v = max(v, D[i].b);
    }
    for (int i = 0; i < nn; ++i) {
      if (D[i].a == n + 1 && D[i].b == n + 1) continue;
      if (D[i].b == n + 1 && D[i].c == n + 1) continue;
      if (D[i].c == n + 1 && D[i].a == n + 1) continue;
      if (D[i].a <= u) continue;
      if (D[i].b <= v) continue;
      if (D[i].c <= w) continue;
      if (D[i].a <= D[i].b && D[i].a <= D[i].c) u = D[i].a;
      if (D[i].b <= D[i].a && D[i].b <= D[i].c) v = D[i].b;
      if (D[i].c <= D[i].b && D[i].c <= D[i].a) w = D[i].c;
    }
    printf("%d\n", u + v + w);
  }
}
