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;
using uint = unsigned;
using lint = long long;
using ulint = unsigned long long;
using ldouble = long double;
using point = complex<double>;
using doublex = complex<ldouble>;
const double E = 1e-9, pi = 2 * acos(0);
const ldouble PI = 2 * acosl(0);
string A, B, C;
int na, nb, nc, nbc, nabc, n, nnabc;
vector<int> memo, digs, nex;
int pack(int a, int b, int c) { return 100 * a + 10 * b + c; }
int pack(int mask, int pos, int posA, int posB, int posC) {
  return mask * n * nabc + pos * nabc + posA * nbc + posB * nc + posC;
}
void unpack(int d, int &a, int &b, int &c) {
  tie(a, b, c) = make_tuple(d / 100, d % 100 / 10, d % 10);
}
void unpack(int state, int &mask, int &pos, int &posA, int &posB, int &posC) {
  tie(mask, pos, posA, posB, posC) =
      make_tuple(state / nnabc, state % nnabc / nabc, state % nabc / nbc,
                 state % nbc / nc, state % nc);
}
int dp(int state) {
  int &res = memo[state];
  if (res != -1) return res;
  res = INT_MAX / 2;
  int mask, pos, posA, posB, posC;
  unpack(state, mask, pos, posA, posB, posC);
  if (!pos) {
    if (mask == 7 and posA == na - 1 and posB == nb - 1 and posC == nc - 1)
      res = 0;
    return res;
  }
  for (int i = 0; i < 10; i++)
    for (int j = 0; j < 10; j++) {
      auto update = [&](bool carry) {
        int k = i + j - (mask & 8) / 8 * 10 + carry;
        if (k < 0 or k >= 10) return;
        int nmask = mask & 7;
        if (carry) nmask |= 8;
        if (i) nmask |= 4;
        if (j) nmask |= 2;
        if (k) nmask |= 1;
        int a = (nmask & 4) ? 1 : 0, b = (nmask & 2) ? 1 : 0,
            c = (nmask & 1) ? 1 : 0;
        int npos = pos - 1;
        int nposA = posA, nposB = posB, nposC = posC;
        if (posA < na - 1 and i == A[posA] - '0') nposA++, a--;
        if (posB < nb - 1 and j == B[posB] - '0') nposB++, b--;
        if (posC < nc - 1 and k == C[posC] - '0') nposC++, c--;
        int nstate = pack(nmask, npos, nposA, nposB, nposC),
            nres = a + b + c + dp(nstate);
        if (nres < res) {
          res = nres;
          nex[state] = nstate;
          digs[state] = pack(i, j, k);
        }
      };
      update(false);
      update(true);
    }
  return res;
}
void get(int state) {
  assert((nex[state] == -1) == (digs[state] == -1));
  if (nex[state] == -1) return;
  int a, b, c;
  unpack(digs[state], a, b, c);
  A += a + '0';
  B += b + '0';
  C += c + '0';
  get(nex[state]);
}
bool solve(int testNumber) {
  string str;
  if (!(cin >> str)) return false;
  int p0 = str.find('+'), p1 = str.find('=');
  A = str.substr(0, p0);
  B = str.substr(p0 + 1, p1 - p0 - 1);
  C = str.substr(p1 + 1);
  na = (int)A.size() + 1;
  nb = (int)B.size() + 1;
  nc = (int)C.size() + 1;
  n = na + nb + nc - 2;
  nbc = nb * nc;
  nabc = na * nbc;
  nnabc = n * nabc;
  memo.assign(nnabc << 4, -1);
  digs = nex = memo;
  int start = pack(0, n - 1, 0, 0, 0);
  dp(start);
  A = B = C = "";
  get(start);
  reverse(A.begin(), A.end());
  reverse(B.begin(), B.end());
  reverse(C.begin(), C.end());
  while (A.back() == '0') A.pop_back();
  while (B.back() == '0') B.pop_back();
  while (C.back() == '0') C.pop_back();
  reverse(A.begin(), A.end());
  reverse(B.begin(), B.end());
  reverse(C.begin(), C.end());
  printf("%s+%s=%s\n", A.c_str(), B.c_str(), C.c_str());
  return true;
}
void init() {}
int main() {
  init();
  for (int i = 1; solve(i); i++)
    ;
  return 0;
}