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.
1
2
3
2
 8
190409007 190409007 352375776 190409007 352375776 352375776 352375776 352375776
190409007 190409007 190409007 190409007 190409007 352375776 352375776 190409007
190409007 352375776 352375776 190409007 190409007 190409007 352375776 352375776
 1
1
1
1
 1
377489979
588153796
588153796
 2
1 1
2 2
3 3
 4
1 1 1 1
1 1 1 1
1 1 1 1
 2
1 2
2 2
1 1
 3
1 1 2
1 1 3
1 1 4

2
 2
 1
 2
 3
 1
 2
 9

#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int a[N], b[N], c[N], d[N * 3], n, m, ans;
int fr1[N * 3], fr2[N * 3], fr3[N * 3];
struct node {
  int l, r, dl, dr, ans, tag;
} t[N * 4];
multiset<int> s;
multiset<int>::iterator ii;
void push_down(int k) {
  if (t[k].tag == -1) return;
  t[k * 2].tag = t[k * 2].dl = t[k * 2].dr = t[k].tag;
  t[k * 2].ans = t[k].tag + t[k * 2].l;
  t[k * 2 + 1].tag = t[k * 2 + 1].dl = t[k * 2 + 1].dr = t[k].tag;
  t[k * 2 + 1].ans = t[k].tag + t[k * 2 + 1].l;
  t[k].tag = -1;
}
void built(int k, int l, int r) {
  t[k].l = l;
  t[k].r = r;
  t[k].tag = -1;
  if (l == r) {
    t[k].dl = t[k].dr = d[l];
    t[k].ans = d[l] + l;
    return;
  }
  int mid = (l + r) >> 1;
  built(k * 2, l, mid);
  built(k * 2 + 1, mid + 1, r);
  t[k].dl = t[k * 2].dl;
  t[k].dr = t[k * 2 + 1].dr;
  t[k].ans = min(t[k * 2].ans, t[k * 2 + 1].ans);
}
void modify(int k, int l, int r, int x) {
  if (t[k].l == l && t[k].r == r && t[k].dl <= x) {
    t[k].dl = t[k].dr = t[k].tag = x;
    t[k].ans = x + t[k].l;
    return;
  }
  if (t[k].l == t[k].r) return;
  push_down(k);
  int mid = (t[k].l + t[k].r) >> 1;
  if (t[k * 2].dr < x && l <= mid) modify(k * 2, l, min(mid, r), x);
  if (t[k * 2 + 1].dr < x && r > mid) modify(k * 2 + 1, max(l, mid + 1), r, x);
  t[k].dl = t[k * 2].dl;
  t[k].dr = t[k * 2 + 1].dr;
  t[k].ans = min(t[k * 2].ans, t[k * 2 + 1].ans);
}
int find(int x) {
  int l = 0, r = m, mid;
  while (r - l > 1) {
    mid = (r + l) >> 1;
    if (d[mid] >= x)
      r = mid;
    else
      l = mid;
  }
  return r;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), d[++m] = a[i];
  for (int i = 1; i <= n; i++) scanf("%d", b + i), d[++m] = b[i];
  for (int i = 1; i <= n; i++) scanf("%d", c + i), d[++m] = c[i];
  sort(d + 1, d + 1 + m);
  m = unique(d + 1, d + 1 + m) - d - 1;
  for (int i = 1; i <= n; i++) a[i] = find(a[i]);
  for (int i = 1; i <= n; i++) b[i] = find(b[i]);
  for (int i = 1; i <= n; i++) c[i] = find(c[i]);
  for (int i = 1; i <= m; i++) fr1[i] = fr2[i] = fr3[i] = n * 3;
  for (int i = 1; i <= n; i++)
    if (fr1[a[i]] == n * 3) fr1[a[i]] = i;
  for (int i = 1; i <= n; i++)
    if (fr2[b[i]] == n * 3) fr2[b[i]] = i;
  for (int i = 1; i <= n; i++)
    if (fr3[c[i]] == n * 3) fr3[c[i]] = i;
  for (int i = 1; i <= m; i++)
    if (fr1[i] == 3 * n) s.insert(fr3[i]);
  s.insert(0);
  for (int i = 0; i <= n; i++) {
    if (i && fr2[b[i]] == i && fr1[b[i]] == 3 * n) {
      ii = s.find(fr3[b[i]]);
      if (ii != s.end()) s.erase(ii);
    }
    ii = s.end();
    ii--;
    d[i] = *ii;
  }
  built(1, 0, n);
  ans = t[1].ans + n;
  for (int i = n; i; i--) {
    if (fr1[a[i]] == i) modify(1, 0, min(n, fr2[a[i]] - 1), fr3[a[i]]);
    ans = min(ans, i - 1 + t[1].ans);
  }
  printf("%d\n", ans);
  return 0;
}
