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:
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>

using namespace std;

#define N 4
#define N2 16
#define LIMIT 45

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

int MDT[N2][N2];

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

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

Puzzle state;
int limit;
int path[LIMIT];
bool dfs(int depth, int prevR) {
    if (state.MD == 0)
        return true;
    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) {
        if (max( prevR,r) - min(r, prevR) == 2)
            continue;
        int tx = sx + dx[r];
        int ty = sy + dy[r];
        if (tx < 0 || ty < 0 || tx >= N || ty >= N)
            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[sx * N + sy], state.f[tx * N + ty]);
        state.space = tx * N + ty;
        if (dfs(depth + 1, r)) {
            path[depth] = r;
            return true;
        }
        state = tmp;
    }
    return false;
}

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

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 p;
    for (int i = 0; i < N2; ++i) {
        scanf("%d", &p.f[i]);
        if (p.f[i] == 0) {
            p.f[i] = N2;
            p.space = i;
        }
    }
    string ans = deepening(p);
    cout << ans.size() << endl;

    return 0;
}