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;
map<int, int> chain, A, F, B, C;
set<int, greater<int> > L;
multiset<int> mn;
void push(int x, int y) {
  map<int, int>::iterator I, J;
  if (chain.count(x)) {
    I = chain.find(x);
  } else {
    I = chain.lower_bound(x);
  }
  int ly = I->second;
  if (ly >= y) return;
  J = I;
  J--;
  while (J->second <= y) {
    mn.erase(mn.find(J->first + ly));
    ly = J->second;
    chain.erase(J);
    J = I;
    J--;
  }
  mn.erase(mn.find(J->first + ly));
  mn.insert(J->first + y);
  if (!chain.count(x)) {
    chain[x] = y;
    mn.insert(x + I->second);
  } else
    I->second = y;
}
int main() {
  chain[0] = 0x0fffffff;
  chain[0x0fffffff] = 0;
  mn.insert(0);
  int N;
  scanf("%d", &N);
  for (int i = 1, x; i <= N; i++) {
    scanf("%d", &x);
    L.insert(x);
    if (!A.count(x)) A[x] = i, F[-i] = x;
  }
  for (int i = 1, x; i <= N; i++) {
    scanf("%d", &x);
    L.insert(x);
    if (!B.count(x)) B[x] = i;
  }
  for (int i = 1, x; i <= N; i++) {
    scanf("%d", &x);
    L.insert(x);
    if (!C.count(x)) C[x] = i;
  }
  for (auto p : L) {
    if (!A.count(p)) {
      int x = B.count(p) ? B[p] : 0x0ffffffe;
      int y = C.count(p) ? C[p] : 0x0ffffffe;
      push(x, y);
    }
  }
  int ans = 3 * N;
  for (auto u : F) {
    int p = u.second;
    if (A.count(p)) {
      if (ans > *mn.begin() + A[p]) ans = *mn.begin() + A[p];
      int x = B.count(p) ? B[p] : 0x0ffffffe;
      int y = C.count(p) ? C[p] : 0x0ffffffe;
      push(x, y);
    }
  }
  if (ans > *mn.begin()) ans = *mn.begin();
  printf("%d\n", ans);
  return 0;
}