DLX解数独。行：81格（每格9种选择）。列：每行每列9格（各9种选择），81格（有或没有）。把已知的先加进去，并把它们标记，之后不再添加。然后把剩下的所有可能都加进去，再dfs求解，应该是唯一解。输出的时候注意一下，先求出真实位置，再求真实数值，最后直接输出字符串。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<queue>
#include<map>
///LOOP
#define REP(i, n) for(int i = 0; i < n; i++)
#define FF(i, a, b) for(int i = a; i < b; i++)
#define FFF(i, a, b) for(int i = a; i <= b; i++)
#define FD(i, a, b) for(int i = a - 1; i >= b; i--)
#define FDD(i, a, b) for(int i = a; i >= b; i--)
///INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RFI(n) scanf("%lf", &n)
#define RFII(n, m) scanf("%lf%lf", &n, &m)
#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
#define RS(s) scanf("%s", s)
///OUTPUT
#define PN printf("\n")
#define PI(n) printf("%d\n", n)
#define PIS(n) printf("%d ", n)
#define PS(s) printf("%s\n", s)
#define PSS(s) printf("%s ", s)
#define PC(n) printf("Case %d: ", n)
///OTHER
#define PB(x) push_back(x)
#define CLR(a, b) memset(a, b, sizeof(a))
#define CPY(a, b) memcpy(a, b, sizeof(b))
#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int MOD = 1e9+7;
const int INFI = 1e9 * 2;
const LL LINFI = 1e17;
const double eps = 1e-6;
const int N = 9;
const int Nr = N * N * N;
const int Nc = N * N * 4;
const int M = Nr * Nc;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};

int U[M], D[M], L[M], R[M], X[M], Y[M], H[Nr], S[Nr], Q[Nr];
bool v[M];
char s[Nr];
int size;

void remove(int c)
{
    L[R[c]] = L[c],R[L[c]] = R[c];
    for(int i = D[c]; i != c; i = D[i])
        for(int j = R[i]; j != i; j = R[j])
            U[D[j]] = U[j], D[U[j]] = D[j], S[Y[j]]--;
}

void resume(int c)
{
    L[R[c]] = R[L[c]] = c;
    for(int i = U[c]; i != c; i = U[i])
        for(int j = L[i]; j != i; j = L[j])
            S[Y[U[D[j]] = D[U[j]] = j]]++;
}

bool dance()
{
    if(!R[0])
    {
        puts(s);
        return 1;
    }
    int p, m = M;
    for(int i = R[0]; i; i = R[i])if(m > S[i])m = S[p = i];
    remove(p);
    for(int i = D[p]; i != p; i = D[i])
    {
        s[(X[i] - 1) / 9] = (X[i] - 1) % 9 + '1';
        for(int j = R[i]; j != i; j = R[j])remove(Y[j]);
        if(dance())return 1;
        for(int j = L[i]; j != i; j = L[j])resume(Y[j]);
    }
    resume(p);
    return 0;
}

void link(int r, int c)
{
    S[Y[size] = c]++;
    D[size] = D[c];
    U[size] = c;
    U[D[c]] = size;
    D[c] = size;
    if(H[r] < 0)H[r] = L[size] = R[size] = size;
    else
    {
        R[size] = R[H[r]];
        L[R[H[r]]] = size;
        L[size] = H[r];
        R[H[r]] = size;
    }
    X[size++] = r;
}

int c[4];

void get(int &r, int i, int j, int k)
{
    r = (i * N + j) * N + k;
    c[0] = i * N + k;
    c[1] = N * N + N * j + k;
    c[2] = N * N * 2 + N * ((i / 3) * 3 + j / 3) + k;
    c[3] = N * N * 3 + i * N + j + 1;
}

void init(int r, int c)
{
    REP(i, c + 1)L[R[i] = i + 1] = U[i] = D[i] = i;
    R[c]=0;
    size = c + 1;
    while(r)H[r--]=-1;
    CLR(S, 0);
    CLR(v, 0);
}

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    int r;
    while(RS(s), s[0] != 'e')
    {
        init(Nr, Nc);
        for(int i = 0, k = 0; i < N; i++)for(int j = 0; j < N; j++, k++)if(s[k] != '.')
        {
            get(r, i, j, s[k] - '0');
            REP(p, 4)link(r, c[p]), v[c[p]] = 1;
        }
        REP(i, 9)REP(j, 9)FFF(k, 1, 9)
        {
            get(r, i, j, k);
            if(v[c[0]] || v[c[1]] || v[c[2]] || v[c[3]])continue;
            REP(p, 4)link(r, c[p]);
        }
        dance();
    }
    return 0;
}