Here we describe a typical problem. There are n balls and n boxes. Each ball is labeled by a unique number from 1 to n. Initially each box contains one of these balls. We can swap two balls in adjacent boxes. We are to sort these balls in increasing order by swaps, i.e. move the ball labeled by 1 to the first box, labeled by 2 to the second box, and so forth. The question is how many swaps are needed.

Now let us consider the situation where the balls are doubled, that is, there are 2n balls and n boxes, exactly two balls are labeled by k for each 1 ≤ k ≤ n, and the boxes contain two balls each. We can swap two balls in adjacent boxes, one ball from each box. We are to move the both balls labeled by 1 to the first box, labeled by 2 to the second box, and so forth. The question is again how many swaps are needed.

Here is one interesting fact. We need 10 swaps to sort [5; 4; 3; 2; 1] (the state with 5 in the first box, 4 in the second box, and so forth): swapping 5 and 4, then 5 and 3, 5 and 2, 5 and 1, 4 and 3, 4 and 2, 4 and 1, 3 and 2, 3 and 1,and finally 2 and 1. Then how many swaps we need to sort [5, 5; 4, 4; 3, 3; 2, 2; 1, 1] (the state with two 5’s in the first box, two 4’s in the second box, and so forth)? Some of you might think 20 swaps - this is not true, but the actual number is 15.

Write a program that calculates the number of swaps for the two-ball version and verify the above fact.



Input

The input consists of multiple datasets. Each dataset has the following format:

n
ball1,1 ball1,2
ball2,1 ball2,2
...
balln,1 balln,2


n is the number of boxes (1 ≤ n ≤ 8). balli,1 and balli,2 , for 1 ≤ i ≤ n, are the labels of two balls initially contained by the i-th box.

The last dataset is followed by a line containing one zero. This line is not a part of any dataset and should not be processed.

Output

For each dataset, print the minumum possible number of swaps.

Example

Input

5
5 5
4 4
3 3
2 2
1 1
5
1 5
3 4
2 5
2 3
1 4
8
8 3
4 2
6 4
3 5
5 8
7 1
2 6
1 7
0


Output

15
9
21
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <math.h>
#include <assert.h>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <map>

using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
static const double EPS = 1e-9;
static const double PI = acos(-1.0);

#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, s, n) for (int i = (s); i < (int)(n); i++)
#define FOREQ(i, s, n) for (int i = (s); i <= (int)(n); i++)
#define FORIT(it, c) for (__typeof((c).begin())it = (c).begin(); it != (c).end(); it++)
#define MEMSET(v, h) memset((v), h, sizeof(v))

int n;
ll encode(int box[2][8]) {
  ll ret = 0;
  REP(lr, 2) {
    REP(index, n) {
      ret |= (ll)box[lr][index] << (3 * (lr * 8 + index));
    }
  }
  return ret;
}
void decode(int box[2][8], ll value) {
  REP(lr, 2) {
    REP(index, n) {
      box[lr][index] = (value >> (3 * (lr * 8 + index))) & 7;
    }
  }
}
int huristic1(int box[2][8]) {
  int ret = 0;
  REP(lr, 2) {
    REP(index, n) {
      ret += abs(box[lr][index] - index);
    }
  }
  return ret;
}
inline int h2check(int l, int r) {
  if (l > r) { return 2; }
  else if (l == r) { return 1; }
  return 0;
}
int huristic2(int box[2][8]) {
  int ret = 5;
  REP(to, n) {
    if (box[0][to] != box[1][to]) { ret++; }
    REP(from, to) {
      REP(fromlr, 2) {
        REP(tolr, 2) {
          ret += h2check(box[fromlr][from], box[tolr][to]);
        }
      }
    }
  }
  return ret;
}
int huristic2part(int box[2][8], int ph2, int f, int t, int flr, int tlr) {
  int ret = ph2;
  if (box[0][f] != box[1][f]) { ret--; }
  if (box[0][t] != box[1][t]) { ret--; }
  ret -= h2check(box[flr][f], box[tlr][t]);
  ret -= h2check(box[flr][f], box[1 ^ tlr][t]);
  ret -= h2check(box[1 ^ flr][f], box[tlr][t]);

  swap(box[flr][f], box[tlr][t]);
  if (box[0][f] != box[1][f]) { ret++; }
  if (box[0][t] != box[1][t]) { ret++; }
  ret += h2check(box[flr][f], box[tlr][t]);
  ret += h2check(box[flr][f], box[1 ^ tlr][t]);
  ret += h2check(box[1 ^ flr][f], box[tlr][t]);
  swap(box[flr][f], box[tlr][t]);
  assert(abs(ret - ph2) <= 6);
  return ret;
}

struct State {
  ll state;
  int cost;
  int hcost;
  int ph2cost;
  State() {;}
  State(ll s, int c, int hc, int ph2) : state(s), cost(c), hcost(hc), ph2cost(ph2) {;}
  bool operator<(const State &rhs) const {
    return cost + hcost > rhs.cost + rhs.hcost;
  }
};
void printBox(int box[2][8]) {
  REP(lr, 2) {
    REP(index, n) {
      printf("%d ", box[lr][index]);
    }
    puts("");
  }
}

const int dy[2] = { 1, -1 };
int box[2][8];

int main() {
  while (scanf("%d", &n) > 0 && n) {
    MEMSET(box, 0);
    REP(i, n) {
      box[0][i] = box[1][i] = i;
    }
    ll endState = encode(box);
    REP(index, n) {
      REP(lr,2) {
        int x;
        scanf("%d", &x);
        x--;
        box[lr][index] = x;
      }
    }
    REP(i, n) {
      if (box[0][i] > box[1][i]) { swap(box[0][i], box[1][i]); }
    }
    set<ll> visit;
    priority_queue<State> que;
    que.push(State(encode(box), 0, 0, huristic2(box)));
    while (!que.empty()) {
      State s = que.top();
      que.pop();
      if (visit.count(s.state)) { continue; }
      visit.insert(s.state);
      if (s.state == endState) {
        printf("%d\n", s.cost);
        break;
      }
      decode(box, s.state);
      REP(from, n - 1) {
        if (box[0][0] == 0 && box[1][0] == 0 && from == 0) { continue; }
        if (box[0][0] == 0 && box[1][0] == 0 && box[0][1] == 1 && box[1][1] == 1 && from == 1) { continue; }
        if (box[0][n - 1] == n - 1 && box[1][n - 1] == n - 1 && from == n - 2) { continue; }
        if (n >= 3 && box[0][n - 1] == n - 1 && box[1][n - 1] == n - 1 && box[0][n - 2] == n - 2 && box[1][n - 2] == n - 2 && from == n - 3) { continue; }
        int to = from + 1;
        REP(fromlr, 2) {
          if (box[0][from] == box[1][from] && fromlr == 1) { continue; }
          REP(tolr, 2) {
            if (box[0][to] == box[1][to] && tolr == 1) { continue; }
            if (box[fromlr][from] == box[tolr][to]) { continue; }
            int h2 = huristic2part(box, s.ph2cost, from, to, fromlr, tolr);
            bool upswap = false;
            bool lowerswap = false;
            swap(box[fromlr][from], box[tolr][to]);
            if (box[0][from] > box[1][from]) {
              swap(box[0][from], box[1][from]);
              upswap = true;
            }
            if (box[0][to] > box[1][to]) {
              swap(box[0][to], box[1][to]);
              lowerswap = true;
            }
            ll enc = encode(box);
            int h1 = huristic1(box);
            //int h2 = huristic2(box);
            que.push(State(enc, s.cost + 1, max(h1 / 2, h2 / 6), h2));
            if (upswap) { swap(box[0][from], box[1][from]); }
            if (lowerswap) { swap(box[0][to], box[1][to]); }
            swap(box[fromlr][from], box[tolr][to]);
          }
        }
      }
    }
   //cout << visit.size() << endl;
  }
}