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
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define INF 9223372036854775807
#define N 4
#define NN 16
#define LIMIT 100

struct Pazzle
{
    int f[NN], space, MD;
};

int MDT[NN][NN];
int limit;
int ans;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
Pazzle state;

int calcAllMD(Pazzle p)
{
    int ret = 0;
    for (int i = 0; i < NN; i++)
    {
        if (p.f[i] == NN)
            continue;
        ret += MDT[i][p.f[i] - 1];
    }
    return ret;
}

bool dfs(int depth, int prev)
{
    if (state.MD == 0)
    {
        ans = depth;
        return true;
    }

    if (state.MD + depth > limit)
        return false;

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

    Pazzle tmp;

    for (int i = 0; i < 4; i++)
    {
        int tx = sx + dx[i];
        int ty = sy + dy[i];

        if (tx < 0 || tx >= N || ty < 0 || ty >= N)
            continue;
        if (abs(prev - i) == 2)
            continue;

        tmp = state;
        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;
        state.MD = calcAllMD(state);
        if (dfs(depth + 1, i))
            return true;
        state = tmp;
    }

    return false;
}

void iterativeDeepening(Pazzle in)
{
    in.MD = calcAllMD(in);

    for (limit = in.MD; limit <= LIMIT; limit++)
    {
        state = in;
        if (dfs(0, -100))
        {
            cout << ans << endl;
            return;
        }
    }
}

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

    Pazzle in;

    for (int i = 0; i < NN; i++)
    {
        cin >> in.f[i];
        if (in.f[i] == 0)
        {
            in.f[i] = NN;
            in.space = i;
        }
    }

    iterativeDeepening(in);
}
