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;
int a[300005], b[300005], c[300005];
int fa[300005], fb[300005], fc[300005];
vector<int> vx;
set<pair<int, int> > st;
set<pair<int, int> >::iterator it, its;
multiset<int> score;
multiset<int>::iterator I;
int n;
void add(int id, int val) {
  bool up = false;
  it = st.lower_bound(pair<int, int>(id + 1, -1));
  if (it != st.end() && (*it).first == id + 1) up = true;
  it--;
  if ((*it).second >= val) return;
  int hi = (*it).second;
  int l;
  while (1) {
    if ((*it).second > val) break;
    l = (*it).first;
    I = score.lower_bound((*it).first + (*it).second);
    score.erase(I);
    if (it == st.begin()) {
      st.erase(it);
      break;
    } else {
      its = it;
      its--;
      st.erase(it);
      it = its;
    }
  }
  st.insert(pair<int, int>(l, val));
  score.insert(l + val);
  if (id + 1 <= n && !up) {
    st.insert(pair<int, int>(id + 1, hi));
    score.insert(id + 1 + hi);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) scanf("%d", &b[i]);
  for (int i = 0; i < n; i++) scanf("%d", &c[i]);
  for (int i = 0; i < n; i++) {
    vx.push_back(a[i]);
    vx.push_back(b[i]);
    vx.push_back(c[i]);
  }
  sort(vx.begin(), vx.end());
  vx.erase(unique(vx.begin(), vx.end()), vx.end());
  for (int i = 0; i < vx.size(); i++) fa[i] = fb[i] = fc[i] = 100000000;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(vx.begin(), vx.end(), a[i]) - vx.begin();
    b[i] = lower_bound(vx.begin(), vx.end(), b[i]) - vx.begin();
    c[i] = lower_bound(vx.begin(), vx.end(), c[i]) - vx.begin();
    if (fa[a[i]] == 100000000) fa[a[i]] = i;
    if (fb[b[i]] == 100000000) fb[b[i]] = i;
    if (fc[c[i]] == 100000000) fc[c[i]] = i;
  }
  st.insert(pair<int, int>(0, 0));
  score.insert(0);
  for (int i = 0; i < vx.size(); i++) {
    if (fa[i] == 100000000) {
      add(fb[i], fc[i] + 1);
    }
  }
  int ret = n + (*score.begin());
  for (int i = n - 1; i >= 0; i--) {
    if (fa[a[i]] == i) add(fb[a[i]], fc[a[i]] + 1);
    ret = min(ret, i + (*score.begin()));
  }
  printf("%d\n", ret);
  return 0;
}
