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;
int codifica(int ia, int ib, int ic, int carry) {
  return ia * 1000 + ib * 100 + ic * 10 + carry;
}
int descia(int codi) { return codi / 1000; }
int descib(int codi) { return (codi / 100) % 10; }
int descic(int codi) { return (codi / 10) % 10; }
int desccarry(int codi) { return codi % 10; }
struct elem {
  int codi;
  int cost;
};
bool operator<(elem e1, elem e2) { return e1.cost > e2.cost; }
vector<int> passaadigits(int x) {
  vector<int> v;
  while (x > 0) {
    v.push_back(x % 10);
    x /= 10;
  }
  return v;
}
struct triple {
  int digit;
  int nexti;
  int cost;
};
vector<triple> computatriple(vector<int> &v, int i) {
  vector<triple> r;
  triple t;
  if (i < int(v.size())) {
    t.digit = v[i];
    t.nexti = i + 1;
    t.cost = 0;
    if (t.nexti < int(v.size())) {
      r.push_back(t);
    } else {
      t.nexti = 8;
      r.push_back(t);
      t.nexti = 9;
      r.push_back(t);
    }
    for (int val = 0; val < 10; val++) {
      t.digit = val;
      t.nexti = i;
      t.cost = 1;
      r.push_back(t);
    }
  } else if (i == 8) {
    t.digit = 0;
    t.nexti = 9;
    t.cost = 0;
    r.push_back(t);
    for (int val = 0; val < 10; val++) {
      t.digit = val;
      t.nexti = i;
      t.cost = 1;
      r.push_back(t);
    }
  } else {
    t.digit = 0;
    t.nexti = 9;
    t.cost = 0;
    r.push_back(t);
  }
  return r;
}
bool codifinal(int codi) {
  int ia, ib, ic, carry;
  ia = descia(codi);
  ib = descib(codi);
  ic = descic(codi);
  carry = desccarry(codi);
  return ia == 9 and ib == 9 and ic == 9 and carry == 0;
}
int codifcodifinal() { return codifica(9, 9, 9, 0); }
int main() {
  int a, b, c;
  char aux1, aux2;
  cin >> a >> aux1 >> b >> aux2 >> c;
  vector<int> va = passaadigits(a);
  vector<int> vb = passaadigits(b);
  vector<int> vc = passaadigits(c);
  elem e;
  int carry = 0;
  int ia = 0;
  int ib = 0;
  int ic = 0;
  e.codi = codifica(ia, ib, ic, carry);
  e.cost = 0;
  priority_queue<elem> q;
  q.push(e);
  vector<int> costcalculat(10000, -1);
  vector<int> anterior(10000);
  vector<int> digita(10000, -1);
  vector<int> digitb(10000, -1);
  vector<int> digitc(10000, -1);
  costcalculat[e.codi] = 0;
  while (not q.empty()) {
    e = q.top();
    q.pop();
    int codi = e.codi;
    int cost = e.cost;
    if (codifinal(codi)) break;
    if (cost == costcalculat[codi]) {
      ia = descia(codi);
      ib = descib(codi);
      ic = descic(codi);
      carry = desccarry(codi);
      vector<triple> ta = computatriple(va, ia);
      vector<triple> tb = computatriple(vb, ib);
      vector<triple> tc = computatriple(vc, ic);
      for (int ita = 0; ita < int(ta.size()); ita++) {
        for (int itb = 0; itb < int(tb.size()); itb++) {
          for (int itc = 0; itc < int(tc.size()); itc++) {
            int da = ta[ita].digit;
            int db = tb[itb].digit;
            int dc = tc[itc].digit;
            int nextia = ta[ita].nexti;
            int nextib = tb[itb].nexti;
            int nextic = tc[itc].nexti;
            int costa = ta[ita].cost;
            int costb = tb[itb].cost;
            int costc = tc[itc].cost;
            if ((da + db + carry) % 10 == dc) {
              int nextcarry = (da + db + carry) / 10;
              int nextcodi = codifica(nextia, nextib, nextic, nextcarry);
              int nextcost = cost + costa + costb + costc;
              if (costcalculat[nextcodi] == -1 or
                  costcalculat[nextcodi] > nextcost) {
                costcalculat[nextcodi] = nextcost;
                anterior[nextcodi] = codi;
                digita[nextcodi] = da;
                digitb[nextcodi] = db;
                digitc[nextcodi] = dc;
                elem nexte;
                nexte.codi = nextcodi;
                nexte.cost = nextcost;
                q.push(nexte);
              }
            }
          }
        }
      }
    }
  }
  int index = codifcodifinal();
  long long int aa = 0;
  long long int bb = 0;
  long long int cc = 0;
  while (index != 0) {
    aa = aa * 10 + digita[index];
    bb = bb * 10 + digitb[index];
    cc = cc * 10 + digitc[index];
    index = anterior[index];
  }
  cout << aa << "+" << bb << "=" << cc << endl;
}