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;
struct triple {
  int a, b, c;
  triple(int a = 0, int b = 0, int c = 0) : a(a), b(b), c(c) {}
};
map<int, triple> M;
int n, res;
int a[110000], b[110000], c[110000];
set<pair<int, int> > cur;
priority_queue<pair<int, int> > Q;
void Insert(int a, int b) {
  set<pair<int, int> >::iterator it = cur.lower_bound(pair<int, int>(a, b));
  if (it != cur.end() && b <= it->second) return;
  if (it != cur.end()) Q.push(pair<int, int>(-(a + it->second), a));
  it = cur.insert(pair<int, int>(a, b)).first;
  while (it != cur.begin()) {
    it--;
    if (it->second <= b)
      cur.erase(it++);
    else {
      it++;
      break;
    }
  }
  if (it != cur.begin()) {
    it--;
    Q.push(pair<int, int>(-(it->first + b), it->first));
  }
}
bool Check(pair<int, int> p) {
  int cand = -p.first, i = p.second;
  set<pair<int, int> >::iterator it = cur.lower_bound(pair<int, int>(i, 0));
  if (it == cur.end() || it->first != i) return false;
  it++;
  if (it == cur.end() || it->second != cand - i) return false;
  return true;
}
int getBest() {
  while (!Q.empty()) {
    pair<int, int> v = Q.top();
    if (Check(v)) return -v.first;
    Q.pop();
  }
  return 1000000000;
}
int main() {
  map<int, triple>::iterator it;
  int i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    it = M.find(a[i]);
    if (it == M.end()) M[a[i]] = triple(i + 1, 1000000000, 1000000000);
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &b[i]);
    it = M.find(b[i]);
    if (it == M.end())
      M[b[i]] = triple(1000000000, i + 1, 1000000000);
    else
      it->second.b = min(it->second.b, i + 1);
  }
  for (i = 0; i < n; i++) {
    scanf("%d", &c[i]);
    it = M.find(c[i]);
    if (it == M.end())
      M[c[i]] = triple(1000000000, 1000000000, i + 1);
    else
      it->second.c = min(it->second.c, i + 1);
  }
  Insert(0, 1000000000);
  Insert(1000000000, 0);
  for (it = M.begin(); it != M.end(); it++) {
    if (it->second.a == 1000000000) Insert(it->second.b, it->second.c);
  }
  res = 1000000000;
  for (int i = n; i >= 0; i--) {
    res = min(res, i + getBest());
    if (i) {
      it = M.find(a[i - 1]);
      if (it->second.a == i) Insert(it->second.b, it->second.c);
    }
  }
  printf("%d\n", res);
  return 0;
}