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;
template <typename T>
using vc = vector<T>;
using pii = pair<long long, long long>;
void xmax(long long& a, long long b) { a = max(a, b); }
void xmin(long long& a, long long b) { a = min(a, b); }
void print(vc<long long>& a, string second) {
  cerr << "prinitng " << second << ": ";
  for (long long i = 0; i < a.size(); i++) {
    cerr << a[i] << " ";
  }
  cerr << endl;
}
bool geht(long long i, long long n) {
  for (long long j = 0; j < 10; j++) {
    if (i & (1 << j)) n--;
  }
  return n == 0;
}
void fill_vec(vc<long long>& cur, const vc<long long>& a, long long k) {
  long long j = 0;
  for (long long i = 0; i < 10; i++) {
    if (k & (1 << i)) cur[i] = a[j++];
  }
}
long long last_left(long long i) {
  long long l = 0;
  for (long long k = 0; k < 10; k++) {
    if (i & (1 << k)) l = k;
  }
  return l;
}
void cut(vc<long long>& a) {
  while (a.back() == 0) a.pop_back();
}
void pprint(vc<long long> a) {
  for (long long i = 0; i < a.size(); i++) {
    cout << a[a.size() - i - 1];
  }
}
long long f_size(vc<long long>& a) {
  long long l = 0;
  for (long long i = 0; i < a.size(); i++) {
    if (a[i] != 0) l = i;
  }
  return l;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  vc<long long> a, b, c;
  string second;
  cin >> second;
  long long index = 0;
  for (long long i = 0; i < second.size(); i++) {
    if (second[i] == '+') {
      index++;
      continue;
    }
    if (second[i] == '=') {
      index++;
      continue;
    }
    switch (index) {
      case 0:
        a.push_back(second[i] - '0');
        break;
      case 1:
        b.push_back(second[i] - '0');
        break;
      case 2:
        c.push_back(second[i] - '0');
    }
  }
  vc<long long> sola;
  vc<long long> solb;
  vc<long long> solc;
  long long length = 100;
  long long n = c.size();
  reverse(a.begin(), a.end());
  reverse(b.begin(), b.end());
  reverse(c.begin(), c.end());
  for (long long i = 0; i < (1 << 10); i++) {
    if (!geht(i, a.size())) continue;
    for (long long j = 0; j < (1 << 10); j++) {
      if (!geht(j, b.size())) continue;
      long long i1 = 0;
      long long i2 = 1;
      if (last_left(i) > last_left(j)) {
        i1 = 1;
        i2 = 0;
      }
      vector<long long> cur[2] = {vc<long long>(10, -1), vc<long long>(10, -1)};
      fill_vec(cur[0], a, i);
      fill_vec(cur[1], b, j);
      long long ind = 0;
      long long mit = 0;
      vc<long long> cur_c(11, 0);
      for (long long k = 0; k < 10; k++) {
        long long sum = 0;
        if (cur[i2][k] == -1) {
          if (cur[i1][k] == -1) cur[i1][k] = 0;
          if (ind >= c.size())
            cur[i2][k] = 0;
          else
            cur[i2][k] = (10 + c[ind] - mit - cur[i1][k]) % 10;
        } else if (cur[i1][k] == -1) {
          if (ind >= c.size())
            cur[i1][k] = 0;
          else
            cur[i1][k] = (10 + c[ind] - mit - cur[i2][k]) % 10;
        }
        if (ind < c.size() && (cur[i1][k] + mit + cur[i2][k]) % 10 == c[ind])
          ind++;
        cur_c[k] = (cur[i1][k] + mit + cur[i2][k]) % 10;
        if (cur[i1][k] + mit + cur[i2][k] >= 10)
          mit = 1;
        else
          mit = 0;
      }
      if (mit == 1) {
        cur_c[10] = 1;
        if (ind > c.size() && c[ind] == 1) ind++;
      }
      if (ind != c.size()) continue;
      long long cur_l = f_size(cur[0]) + f_size(cur[1]) + f_size(cur_c);
      if (length > cur_l) {
        length = cur_l;
        sola = cur[0];
        solb = cur[1];
        solc = cur_c;
      }
    }
  }
  cut(sola);
  cut(solb);
  cut(solc);
  pprint(sola);
  cout << "+";
  pprint(solb);
  cout << "=";
  pprint(solc);
  cout << "\n";
}