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 <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
#define ll long long
#define endl '\n'
using namespace std;
#define N 4
#define N2 16
#define LIMIT 100

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

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

Puzzle state;
int limit;  // 深さの限界
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 || 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() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    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;
}