Problem: The goal of the 15 puzzle problem is to complete pieces on $4 \times 4$ cells where one of the cells is empty space.

In this problem, the space is represented by 0 and pieces are represented by integers from 1 to 15 as shown below.


1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15


You can move a piece toward the empty space at one step. Your goal is to make the pieces the following configuration in the shortest move (fewest steps).


1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 0


Write a program which reads an initial state of the puzzle and prints the fewest steps to solve the puzzle.

Constraints

* The given puzzle is solvable in at most 45 steps.

Input

The $4 \times 4$ integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Example

Input

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15


Output

8

Solution (in CPP):

Solution:
// Iterative Deepening
#include<stdio.h>
#include<iostream>
#include<cmath>
#include<string>
#include<cassert>
using namespace std;
#define N 4
#define N2 16
#define LIMIT 100

static const int dx[4] = {0, -1, 0, 1};
static const int dy[4] = {1, 0, -1, 0};
static const char dir[4] = {'r', 'u', 'l', 'd'};
int MDT[N2][N2];

struct Puzzle { int f[N2], space, MD; };

Puzzle state;
int limit;  /* 深さの制限 */
int path[LIMIT];

int getAllMD(Puzzle pz) {
  int sum = 0;
  for ( int i = 0; i < N2; i++ ) {
    if ( pz.f[i] == N2 ) continue;
    sum += MDT[i][pz.f[i] - 1];
  }
  return sum;
}

bool isSolved() {
  for ( int i = 0; i < N2; i++ ) if ( state.f[i] != i + 1 ) return false;
  return true;
}


bool dfs(int depth, int prev) {
  if ( state.MD == 0 ) return true;
  /* 現在の深さにヒューリスティックを足して制限を超えたら枝を刈る */
  if ( depth + state.MD > limit ) return false;

  int sx = state.space / N;
  int sy = state.space % N;
  Puzzle tmp;

  for ( int r = 0; r < 4; r++ ) {
    int tx = sx + dx[r];
    int ty = sy + dy[r];
    if ( tx < 0 || ty < 0 || tx >= N || ty >= N ) continue;
    if ( max(prev, r)-min(prev, r) == 2 ) continue;
    tmp = state;
    /* マンハッタン距離の差分を計算しつつ、ピースをスワップ */
    state.MD -= MDT[tx * N + ty][state.f[tx * N + ty] - 1];
    state.MD += MDT[sx * N + sy][state.f[tx * N + ty] - 1];
    swap(state.f[tx * N + ty], state.f[sx * N + sy]);
    state.space = tx * N + ty;
    if ( dfs(depth + 1, r) ) { path[depth] = r; return true; }
    state = tmp;
  }

  return false;
}

/* 反復深化 */
string iterative_deepening(Puzzle in) {
  in.MD = getAllMD(in);  /* 初期状態のマンハッタン距離 */

  for ( limit = in.MD; limit <= LIMIT; limit++ ) {
    state = in;
    if ( dfs(0, -100) ) {
      string ans = "";
      for ( int i = 0; i < limit; i++ ) ans += dir[path[i]];
      return ans;
    }
  }

  return "unsolvable";
}

int main() {
  for ( int i = 0; i < N2; i++ )
    for ( int j = 0; j < N2; j++ )
      MDT[i][j] = abs(i / N - j / N) + abs(i % N - j % N);

  Puzzle in;

  for ( int i = 0; i < N2; i++ ) {
    cin >> in.f[i];
    if ( in.f[i] == 0 ) {
      in.f[i] = N2;
      in.space = i;
    }
  }
  string ans = iterative_deepening(in);
  cout << ans.size() << endl;

  return 0;
}