#include <bits/stdc++.h>
using namespace std;

typedef unordered_map<string, int> MSI;
typedef unordered_map<string, pair<char, string>> MSP;

string st, ed = "12345678x";

//上下左右
char op[] = {'u', 'd', 'l', 'r'};
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};

//前驱
MSP pre;

//伟大的结构体放光芒
struct Node {
    int dist;
    string str;
    bool operator<(const Node &t) const {
        return dist > t.dist; //距离大的反而小，距离小的反而大。所以，配合优先队列（大顶堆），小的在上，大的在下
    }
};

//估值函数,用曼哈顿距离估值
int f(string str) {
    int res = 0;
    for (int i = 0; i < str.size(); i++)
        if (str[i] != 'x') {
            int t = str[i] - '1';
            res += abs(i / 3 - t / 3) + abs(i % 3 - t % 3);
        }
    return res;
}

string astar() {
    MSI dist;
    priority_queue<Node> q;

    //将估价函数与字符串形成数对，放入到优先队列中
    q.push({f(st), st});
    dist[st] = 0; //记录距离为0

    while (q.size()) {
        string t = q.top().str;
        q.pop();

        if (t == ed) break;

        int k = t.find('x');
        int tx = k / 3, ty = k % 3;
        for (int i = 0; i < 4; i++) {
            int x = tx + dx[i], y = ty + dy[i];
            if (x < 0 || x >= 3 || y < 0 || y >= 3) continue; //出界
            string v = t;                                     //将t字符串复制出来一个，生成 v字符串
            swap(v[k], v[x * 3 + y]);                         //将原字符串中x与目标位置进行交换，生成新的目标状态字符串v

            if (dist[v]) continue;
            dist[v] = dist[t] + 1;       // 更新最小值
            q.push({dist[v] + f(v), v}); // 将新的估价函数计算更新，重新入队列
            pre[v] = {op[i], t};
        }
    }

    //将路径倒转过来
    string res;
    while (ed != st) {
        res += pre[ed].first;
        ed = pre[ed].second;
    }
    reverse(res.begin(), res.end());
    return res;
}

int main() {
    char c;
    for (int i = 0; i < 9; i++) cin >> c, st += c;

    //八数码定理
    int nums = 0;
    for (int i = 0; i < 9; i++) {
        if (st[i] == 'x') continue;
        for (int j = i + 1; j < 9; j++) {
            if (st[j] == 'x') continue;
            if (st[j] < st[i]) nums++; //逆序数
        }
    }
    //奇数个逆序对，无解
    if (nums & 1)
        puts("unsolvable");
    else //运行astar算法，返回最短距离
        cout << astar() << endl;

    return 0;
}