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;
int64_t a, b, c;
string sa, sb, sc;
constexpr int N = 19;
int f[N][N][N][2];
tuple<int64_t, int64_t, int64_t> g[N][N][N][2];
int main() {
  ios_base::sync_with_stdio(false);
  char x;
  cin >> a >> x >> b >> x >> c;
  sa = "0" + to_string(a);
  sb = "0" + to_string(b);
  sc = "0" + to_string(c);
  int na = sa.length(), nb = sb.length(), nc = sc.length();
  memset(f, 0x3f, sizeof(f));
  f[0][0][0][0] = 0;
  g[0][0][0][0] = {0, 0, 0};
  int64_t p10[N];
  p10[0] = 1;
  for (int i = 1; i < N; ++i) p10[i] = p10[i - 1] * 10;
  for (int i = 0; i <= na; ++i)
    for (int j = 0; j <= nb; ++j)
      for (int k = 0; k <= nc; ++k)
        for (int r : {0, 1}) {
          int ff = f[i][j][k][r];
          if (ff >= N) continue;
          int w;
          tie(a, b, w) = g[i][j][k][r];
          int64_t pw = p10[w];
          c = a + b;
          int64_t pa = pw * (i == na ? 0 : sa[na - 1 - i] - '0');
          int64_t pb = pw * (j == nb ? 0 : sb[nb - 1 - j] - '0');
          int64_t pc = pw * (k == nc ? 0 : sc[nc - 1 - k] - '0');
          int64_t pr = pw * r;
          if (k == nc - 1 && i >= na - 1 && j >= nb - 1) {
            if (ff + r < f[na][nb][k + 1][0]) {
              f[na][nb][k + 1][0] = ff + r;
              g[na][nb][k + 1][0] = {a, b, w};
            }
          }
          if (i < na && j < nb && k < nc) {
            int nr = 0;
            for (int64_t t : {c + pc, c + pc + pw * 10}) {
              if (a + pa + b + pb + pr == t &&
                  ff < f[i + 1][j + 1][k + 1][nr]) {
                f[i + 1][j + 1][k + 1][nr] = ff;
                g[i + 1][j + 1][k + 1][nr] = {a + pa, b + pb, w + 1};
              }
              ++nr;
            }
          }
          if (i < na && j < nb) {
            int nr = 0;
            for (int64_t t : {pr + pa + pb, pr + pa + pb - pw * 10}) {
              if (t % pw == 0 && t >= 0 && t / pw <= (k == nc ? 0 : 9) &&
                  ff + (t > 0 || k < nc) < f[i + 1][j + 1][k][nr]) {
                f[i + 1][j + 1][k][nr] = ff + (t > 0 || k < nc);
                g[i + 1][j + 1][k][nr] = {a + pa, b + pb, w + 1};
              }
              ++nr;
            }
          }
          if (k < nc) {
            int nr = 0;
            for (int64_t t : {pc - pa - pr, pc + pw * 10 - pa - pr}) {
              if (t % pw == 0 && t / pw <= (j == nb ? 0 : 9) && t >= 0 &&
                  ff + (t > 0 || j < nb) < f[min(na, i + 1)][j][k + 1][nr]) {
                f[min(na, i + 1)][j][k + 1][nr] = ff + (t > 0 || j < nb);
                g[min(na, i + 1)][j][k + 1][nr] = {a + pa, b + t, w + 1};
              }
              ++nr;
            }
          }
          if (k < nc) {
            int nr = 0;
            for (int64_t t : {pc - pb - pr, pc + pw * 10 - pb - pr}) {
              if (t % pw == 0 && t / pw <= (i == na ? 0 : 9) && t >= 0 &&
                  ff + (t > 0 || i < na) < f[i][min(nb, j + 1)][k + 1][nr]) {
                f[i][min(nb, j + 1)][k + 1][nr] = ff + (t > 0 || i < na);
                g[i][min(nb, j + 1)][k + 1][nr] = {a + t, b + pb, w + 1};
              }
              ++nr;
            }
          }
          if (i < na && j == nb) {
            int nr = 0;
            for (int64_t t : {pa + pr, pa + pr - pw * 10}) {
              if (t % pw == 0 && t / pw <= (k == nc ? 0 : 9) && t >= 0 &&
                  ff + (t > 0 || k < nc) < f[i + 1][j][k][nr]) {
                f[i + 1][j][k][nr] = ff + (t > 0 || k < nc);
                g[i + 1][j][k][nr] = {a + pa, b, w + 1};
              }
              ++nr;
            }
          }
          if (i == na && j < nb) {
            int nr = 0;
            for (int64_t t : {pb + pr, pb + pr - pw * 10}) {
              if (t % pw == 0 && t / pw <= (k == nc ? 0 : 9) && t >= 0 &&
                  ff + (t > 0 || k < nc) < f[i][j + 1][k][nr]) {
                f[i][j + 1][k][nr] = ff + (t > 0 || k < nc);
                g[i][j + 1][k][nr] = {a, b + pb, w + 1};
              }
              ++nr;
            }
          }
        }
  tie(a, b, c) = g[na][nb][nc][0];
  cout << a << "+" << b << "=" << a + b << endl;
  return 0;
}