Problem: One day Vasya was solving arithmetical problems. He wrote down an expression a + b = c in his notebook. When the teacher checked Vasya's work it turned out that Vasya had solved the problem incorrectly. Now Vasya tries to find excuses. He says that he simply forgot to write down several digits in numbers a, b and c, but he can't remember what numbers they actually were. Help Vasya, find such numbers x, y and z, with which the following conditions are met: 

  * x + y = z, 
  * from the expression x + y = z several digits can be erased in such a way that the result will be a + b = c, 
  * the expression x + y = z should have the minimal length. 

Input

The first and only input line contains the expression a + b = c (1 ≤ a, b, c ≤ 106, a, b and c don't contain leading zeroes) which is the expression Vasya wrote down.

Output

Print the correct expression x + y = z (x, y and z are non-negative numbers without leading zeroes). The expression a + b = c must be met in x + y = z as a subsequence. The printed solution should have the minimal possible number of characters. If there are several such solutions, you can print any of them.

Examples

Input

2+4=5


Output

21+4=25


Input

1+1=3


Output

1+31=32


Input

1+1=2


Output

1+1=2

Solution (in JAVA):

Solution:
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.Scanner;

public class Main {
    private static final int oo = 0x3F3F3F3F;
    private static Scanner cin;
    private static int[] a;
    private static int[] b;
    private static int[] c;
    private static int[][][][] dp;
    private static String[][][][][] ret;

    public static void main(String[] args) throws IOException {
        File fin = new File("fish_in.txt");
        if (fin.exists()) {
            System.setIn(new FileInputStream(fin));
        }
        cin = new Scanner(System.in);

        while (cin.hasNext()) {
            init();
            dp = new int[a.length + 1][b.length + 1][c.length][2];
            ret = new String[a.length + 1][b.length + 1][c.length][2][2];

            for (int i = 0; i <= a.length; i++)
                for (int j = 0; j <= b.length; j++)
                    for (int k = 0; k < c.length; k++)
                        for (int fd = 0; fd < 2; fd++) {
                            dp[i][j][k][fd] = oo;
                            ret[i][j][k][fd][0] = ret[i][j][k][fd][1] = "";
                        }

            dp[0][0][0][0] = 0;

            for (int i = 0; i <= a.length; i++) {
                for (int j = 0; j <= b.length; j++) {
                    for (int k = 0; k < c.length; k++) {
                        for (int fd = 0; fd < 2; fd++) {
                            if (dp[i][j][k][fd] == oo)
                                continue;
                            if (i == a.length - 1)
                                doit(i + 1, j, k, fd, dp[i][j][k][fd], ret[i][j][k][fd]);
                            if (j == b.length - 1)
                                doit(i, j + 1, k, fd, dp[i][j][k][fd], ret[i][j][k][fd]);

                            for (int di = 0; di < (i < a.length ? 10 : 1); di++) {
                                for (int dj = 0; dj < (j < b.length ? 10 : 1); dj++) {
                                    int ni = i + (i < a.length && di == a[i] ? 1 : 0);
                                    int nj = j + (j < b.length && dj == b[j] ? 1 : 0);
                                    int nk = k + ((di + dj + fd) % 10 == c[k] ? 1 : 0);
                                    int nf = (di + dj + fd) / 10;
                                    int nv = dp[i][j][k][fd] + (i < a.length ? 1 : 0) + (j < b.length ? 1 : 0) + 1;
                                    if (dp[ni][nj][nk][nf] > nv&&!(ni==i&&nj==j&&nk==k)) {
                                        dp[ni][nj][nk][nf] = nv;
                                        ret[ni][nj][nk][nf][0] = ret[i][j][k][fd][0];
                                        ret[ni][nj][nk][nf][1] = ret[i][j][k][fd][1];
                                        if (i < a.length)
                                            ret[ni][nj][nk][nf][0] += di;
                                        if (j < b.length)
                                            ret[ni][nj][nk][nf][1] += dj;
                                    }
                                }
                            }
                        }
                    }
                }
            }

            String[] ans = ret[a.length][b.length][c.length - 1][0];
            long A = 0, B = 0;
            for (int i = ans[0].length() - 1; i >= 0; i--) {
                A = A * 10 + ans[0].charAt(i) - '0';
            }
            for (int i = ans[1].length() - 1; i >= 0; i--) {
                B = B * 10 + ans[1].charAt(i) - '0';
            }
            System.out.println(A + "+" + B + "=" + (A + B));
        }
    }

    private static void doit(int i, int j, int k, int fd, int v, String[] s) {
        // TODO Auto-generated method stub
        if (dp[i][j][k][fd] > v) {
            dp[i][j][k][fd] = v;
            ret[i][j][k][fd][0] = s[0];
            ret[i][j][k][fd][1] = s[1];
        }
    }

    private static void init() {
        // TODO Auto-generated method stub
        String buffer = cin.nextLine();
        String[] tmp = new String[3];
        tmp[0] = tmp[1] = tmp[2] = "";
        for (int i = 0, j = 0; i < buffer.length(); i++) {
            if (buffer.charAt(i) == '+' || buffer.charAt(i) == '=') {
                j++;
            } else {
                tmp[j] += buffer.charAt(i);
            }
        }
        a = doit(tmp[0]);
        b = doit(tmp[1]);
        c = doit(tmp[2]);
    }

    private static int[] doit(String s) {
        // TODO Auto-generated method stub
        int[] ret = new int[s.length() + 1];
        for (int i = 0; i < s.length(); i++) {
            ret[i] = s.charAt(s.length() - 1 - i) - '0';
        }
        ret[s.length()] = -1;
        return ret;
    }
}