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 double eps = 1e-9;
const int INF = 1 << 30;
const long long LINF = 1ll << 60;
const int BUFFER_SZ = 1 << 15;
const int MOD = 1000 * 1000 * 1000 + 7;
char BUFFER[BUFFER_SZ];
int gi() {
  int x;
  scanf("%d", &x);
  return x;
}
double gd() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long gll() {
  long long x;
  cin >> x;
  return x;
}
vector<int> gvi(int n) {
  vector<int> a;
  while (n-- > 0) a.push_back(gi());
  return a;
}
string gs() {
  scanf("%s", BUFFER);
  return string(BUFFER);
}
template <class T>
void print(vector<T> &x, string format) {
  format += "%c";
  for (int i = 0; i < x.size(); ++i)
    printf(format.c_str(), x[i], i + 1 == x.size() ? '\n' : ' ');
}
int dp[50][10][10][10][2][2][2][2];
vector<int> dpP[50][10][10][10][2][2][2][2];
vector<int> a, b, c;
int go(int pos, int p1, int p2, int p3, int nz1, int nz2, int nz3, int carry) {
  int &r = dp[pos][p1][p2][p3][nz1][nz2][nz3][carry];
  if (r == -1) {
    r = 1 << 20;
    if (pos > 40) return r;
    if (!carry && a.size() == p1 && b.size() == p2 && c.size() == p3)
      return r = 0;
    for (int d1 = 0; d1 < 10; ++d1) {
      for (int d2 = 0; d2 < 10; ++d2) {
        for (int cr = 0; cr < 2; ++cr) {
          if (carry && d1 + d2 + cr < 10) continue;
          if (!carry && d1 + d2 + cr >= 10) continue;
          int pp1 = p1;
          if (p1 < a.size() && a[p1] == d1) ++pp1;
          int pp2 = p2;
          if (p2 < b.size() && b[p2] == d2) ++pp2;
          int v = (d1 + d2 + cr) % 10;
          int pp3 = p3;
          if (p3 < c.size() && c[p3] == v) ++pp3;
          int nnz1 = (d1 > 0 || nz1) ? 1 : 0;
          int nnz2 = (d2 > 0 || nz2) ? 1 : 0;
          int nnz3 = (v > 0 || nz3) ? 1 : 0;
          int rr = go(pos + 1, pp1, pp2, pp3, nnz1, nnz2, nnz3, cr) + nnz1 +
                   nnz2 + nnz3;
          if (rr < r) {
            dpP[pos][p1][p2][p3][nz1][nz2][nz3][carry] = {
                d1, d2, v, pp1, pp2, pp3, nnz1, nnz2, nnz3, cr};
            r = rr;
          }
        }
      }
    }
  }
  return r;
}
int main() {
  string s = gs();
  int p = 0;
  while (s[p] != '+') a.push_back(s[p++] - '0');
  ++p;
  while (s[p] != '=') b.push_back(s[p++] - '0');
  ++p;
  while (p < s.size()) c.push_back(s[p++] - '0');
  memset((dp), (-1), sizeof(dp));
  int res = go(0, 0, 0, 0, 0, 0, 0, 0);
  int pos = 0;
  int p1 = 0;
  int p2 = 0;
  int p3 = 0;
  int nz1 = 0;
  int nz2 = 0;
  int nz3 = 0;
  int carry = 0;
  vector<int> A, B, C;
  while (p1 < a.size() || p2 < b.size() || p3 < c.size()) {
    vector<int> move = dpP[pos][p1][p2][p3][nz1][nz2][nz3][carry];
    if (A.size() || move[0]) A.push_back(move[0]);
    if (B.size() || move[1]) B.push_back(move[1]);
    if (C.size() || move[2]) C.push_back(move[2]);
    p1 = move[3];
    p2 = move[4];
    p3 = move[5];
    nz1 = move[6];
    nz2 = move[7];
    nz3 = move[8];
    carry = move[9];
    ++pos;
  }
  for (int x : A) printf("%d", x);
  printf("+");
  for (int x : B) printf("%d", x);
  printf("=");
  for (int x : C) printf("%d", x);
  printf("\n");
  return 0;
}