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 INF = 0x3f3f3f3f;
int n, tot = 0, fa[3 * 100005], fb[3 * 100005], fc[3 * 100005], a[100005],
       b[100005], c[100005];
int ans = INF, flag = 1;
struct node {
  int val, id1, id2;
  friend bool operator<(node a, node b) { return a.val < b.val; }
} f[3 * 100005];
struct Pair {
  int b, c;
  friend bool operator<(Pair a, Pair b) { return a.b < b.b; }
} p;
set<Pair> s;
set<Pair>::iterator it;
multiset<int> mul_s;
multiset<int>::iterator mul_it;
inline int read() {
  register int x = 0, f = 1;
  register char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
inline void insertPair(int newb, int newc) {
  int nowb, nowc;
  Pair e = {newb, newc};
  it = s.upper_bound(e);
  if ((*it).c >= e.c) return;
  nowc = (*it).c;
  it--;
  mul_it = mul_s.find((*it).b + nowc);
  mul_s.erase(mul_it);
  while ((*it).c <= e.c) {
    nowb = (*it).b, nowc = (*it).c;
    it--;
    mul_s.erase((*it).b + nowc);
    it++;
    s.erase(it);
    it = s.lower_bound(e);
    it--;
  }
  nowb = (*it).b, nowc = (*it).c;
  mul_s.insert(nowb + e.c);
  it++;
  nowc = (*it).c;
  mul_s.insert(e.b + nowc);
  s.insert(e);
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    f[++tot].val = read();
    f[tot].id1 = 1, f[tot].id2 = i;
  }
  for (int i = 1; i <= n; i++) {
    f[++tot].val = read();
    f[tot].id1 = 2, f[tot].id2 = i;
  }
  for (int i = 1; i <= n; i++) {
    f[++tot].val = read();
    f[tot].id1 = 3, f[tot].id2 = i;
  }
  sort(f + 1, f + 3 * n + 1);
  tot = 0;
  for (int i = 1; i <= 3 * n; i++) {
    if (i == 1 || f[i].val != f[i - 1].val) tot++;
    if (f[i].id1 == 1) a[f[i].id2] = tot;
    if (f[i].id1 == 2) b[f[i].id2] = tot;
    if (f[i].id1 == 3) c[f[i].id2] = tot;
  }
  for (int i = 1; i <= n; i++)
    if (!fa[a[i]]) fa[a[i]] = i;
  for (int i = 1; i <= n; i++)
    if (!fb[b[i]]) fb[b[i]] = i;
  for (int i = 1; i <= n; i++)
    if (!fc[c[i]]) fc[c[i]] = i;
  p.b = 0, p.c = INF + 1;
  s.insert(p);
  p.b = INF + 1, p.c = 0;
  s.insert(p);
  mul_s.insert(0);
  for (int i = 1; i <= tot; i++)
    if (!fa[i]) {
      if (fb[i]) {
        if (fc[i])
          insertPair(fb[i], fc[i]);
        else
          insertPair(fb[i], INF);
      } else {
        if (fc[i])
          insertPair(INF, fc[i]);
        else
          insertPair(INF, INF);
      }
    }
  for (int i = n; i >= 1; i--) {
    ans = min(ans, i + (*mul_s.begin()));
    if (fa[a[i]] == i) {
      if (!fb[a[i]] && !fc[a[i]]) {
        flag = 0;
        break;
      }
      if (fb[a[i]]) {
        if (fc[a[i]])
          insertPair(fb[a[i]], fc[a[i]]);
        else
          insertPair(fb[a[i]], INF);
      } else {
        if (fc[a[i]])
          insertPair(INF, fc[a[i]]);
        else
          insertPair(INF, INF);
      }
    }
  }
  if (flag) ans = min(ans, (*mul_s.begin()));
  printf("%d", ans);
  return 0;
}
