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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n), b(n), c(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) cin >> b[i];
  for (int i = 0; i < n; ++i) cin >> c[i];
  for (auto k : c) {
    b.push_back(k);
  }
  map<int, int> whereb;
  map<int, int> wherec;
  for (int i = (int)b.size() - 1; i >= 0; --i) whereb[b[i]] = i;
  for (int i = (int)c.size() - 1; i >= 0; --i) wherec[c[i]] = i;
  vector<bool> first_in_a(a.size(), false);
  vector<bool> first_in_b(b.size(), false);
  set<int> ina, inb;
  for (int i = 0; i < a.size(); ++i) {
    if (!ina.count(a[i])) {
      ina.insert(a[i]);
      first_in_a[i] = true;
    }
  }
  for (int i = 0; i < b.size(); ++i) {
    if (!inb.count(b[i])) {
      inb.insert(b[i]);
      first_in_b[i] = true;
    }
  }
  set<pair<int, int>> maxs;
  multiset<int> res;
  maxs.emplace(1e9, 0);
  maxs.emplace(-1, 1e9 + 5);
  res.insert(maxs.begin()->first + next(maxs.begin())->second + 1);
  auto del = [&](pair<int, int> p) {
    auto it = maxs.find(p);
    assert(it != maxs.end());
    auto inext = next(it);
    auto iprev = prev(it);
    res.erase(res.find(iprev->first + it->second + 1));
    res.erase(res.find(it->first + inext->second + 1));
    maxs.erase(it);
    res.insert(iprev->first + inext->second + 1);
  };
  auto add = [&](pair<int, int> p) {
    auto it = maxs.lower_bound(make_pair(p.first, -5));
    if (it->second >= p.second) return;
    if (it->first == p.first) {
      ++it;
    }
    while (prev(it)->second <= p.second) del(*prev(it));
    maxs.insert(p);
    it = maxs.find(p);
    auto inext = next(it);
    auto iprev = prev(it);
    res.insert(iprev->first + it->second + 1);
    res.insert(it->first + inext->second + 1);
    res.erase(res.find(iprev->first + inext->second + 1));
  };
  for (int i = 0; i < b.size(); ++i) {
    if (first_in_b[i] && !ina.count(b[i])) {
      int inc = 1e9;
      if (wherec.count(b[i])) inc = wherec[b[i]] + 1;
      add({i, inc});
    }
  }
  int ans = n + *res.begin();
  for (int i = n - 1; i >= 0; --i) {
    if (first_in_a[i]) {
      if (!inb.count(a[i])) break;
      int inc = 1e9;
      if (wherec.count(a[i])) inc = wherec[a[i]] + 1;
      add({whereb[a[i]], inc});
    }
    ans = min(ans, i + *res.begin());
  }
  cout << ans << '\n';
  return 0;
}