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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 16;
const int inf = 1e9 + 7;
int f[N][N][N][2][2][2];
struct node {
  int x;
  int y;
  int z;
  int i1;
  int j1;
  int k1;
  int p1;
  int fi1, fi2;
} pre[N][N][N][2][2][2];
int a[N], b[N], c[N];
char s[N * 5];
inline void Upd(int &x, int y, node &v, int t1, int t2, int t3, int ii, int jj,
                int kk, int pp, int p1, int p2) {
  if (x == -1) x = inf;
  if (x <= y) return;
  x = y;
  v.x = t1;
  v.y = t2;
  v.z = t3;
  v.i1 = ii;
  v.j1 = jj;
  v.k1 = kk;
  v.p1 = pp;
  v.fi1 = p1;
  v.fi2 = p2;
}
int main() {
  scanf("%s", s + 1);
  int na, nb, nc;
  na = nb = nc = 0;
  memset(a, -1, sizeof(a));
  memset(b, -1, sizeof(b));
  memset(c, -1, sizeof(c));
  int len = strlen(s + 1);
  int ii = 1;
  while (s[ii] != '+') {
    a[++na] = s[ii] - '0';
    ii++;
  }
  ii++;
  while (s[ii] != '=') {
    b[++nb] = s[ii] - '0';
    ii++;
  }
  ii++;
  while (ii <= len) {
    c[++nc] = s[ii] - '0';
    ii++;
  }
  reverse(a + 1, a + 1 + na);
  reverse(b + 1, b + 1 + nb);
  reverse(c + 1, c + 1 + nc);
  memset(f, -1, sizeof(f));
  f[0][0][0][0][0][0] = 0;
  for (int i = (0); i <= (na); i++)
    for (int j = (0); j <= (nb); j++)
      for (int k = (0); k <= (nc); k++)
        for (int up = (1); up >= (0); up--)
          for (int p1 = (0); p1 <= (1); p1++)
            for (int p2 = (0); p2 <= (1); p2++)
              if (f[i][j][k][up][p1][p2] != -1) {
                for (int t1 = (0); t1 <= (9); t1++)
                  for (int t2 = (0); t2 <= (9); t2++) {
                    int t3 = t1 + t2 + up;
                    for (int q1 = (p1); q1 <= (1); q1++)
                      for (int q2 = (p2); q2 <= (1); q2++) {
                        int nxt = f[i][j][k][up][p1][p2] + 3;
                        if (p1 && (i < na || t1)) continue;
                        if (p2 && (i < na || t2)) continue;
                        if ((i == na && !t1) && q1) nxt--;
                        if ((j == nb && !t2) && q2) nxt--;
                        Upd(f[i + (t1 == a[i + 1])][j + (t2 == b[j + 1])]
                             [k + ((t3 % 10) == c[k + 1])][t3 / 10][q1][q2],
                            nxt,
                            pre[i + (t1 == a[i + 1])][j + (t2 == b[j + 1])]
                               [k + ((t3 % 10) == c[k + 1])][t3 / 10][q1][q2],
                            t1, t2, t3 % 10, i, j, k, up, p1, p2);
                      }
                  }
              }
  int i = na, j = nb, k = nc, p = 0, t1 = 1, t2 = 1;
  long long ra = 0, rb = 0, rc = 0;
  while (i || j || k) {
    ra = 10ll * ra + pre[i][j][k][p][t1][t2].x;
    rb = 10ll * rb + pre[i][j][k][p][t1][t2].y;
    rc = 10ll * rc + pre[i][j][k][p][t1][t2].z;
    int ti = i, tj = j, tk = k, tp = p, n1 = t1, n2 = t2;
    i = pre[ti][tj][tk][tp][n1][n2].i1;
    j = pre[ti][tj][tk][tp][n1][n2].j1;
    k = pre[ti][tj][tk][tp][n1][n2].k1;
    p = pre[ti][tj][tk][tp][n1][n2].p1;
    t1 = pre[ti][tj][tk][tp][n1][n2].fi1;
    t2 = pre[ti][tj][tk][tp][n1][n2].fi2;
  }
  printf("%lld+%lld=%lld\n", ra, rb, rc);
  return 0;
}