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;
char A[10], B[10], C[10];
int best[11][11][11][11][11][11][2];
int a, b, c;
string ret1, ret2, ret3;
int solve(int ind1, int ind2, int ind3, int l1, int l2, int l3, bool carry) {
  if (l1 > 10 || l2 > 10 || l3 > 10) return 1 << 30;
  if (ind1 == 0 && ind2 == 0 && ind3 == 0) {
    if (carry) return 1;
    return -(l1 + l2 + l3);
  }
  if (best[ind1][ind2][ind3][l1][l2][l3][carry] != -1)
    return best[ind1][ind2][ind3][l1][l2][l3][carry];
  int ret = 1 << 28;
  int i, j, k;
  if (ind1 > 0) {
    i = A[ind1] - '0';
    for (j = 0; j < 10; j++) {
      k = (i + j + carry) % 10;
      int toAdd = 0, nind1 = ind1, nind2 = ind2, nind3 = ind3,
          ncarry = (i + j + carry > 9);
      if (ind1 > 0 && i == A[ind1] - '0')
        nind1--;
      else
        toAdd++;
      if (ind2 > 0 && j == B[ind2] - '0')
        nind2--;
      else
        toAdd++;
      if (ind3 > 0 && k == C[ind3] - '0')
        nind3--;
      else
        toAdd++;
      ret = min(ret, solve(nind1, nind2, nind3, (i ? 0 : (l1 + 1)),
                           (j ? 0 : (l2 + 1)), (k ? 0 : (l3 + 1)), ncarry) +
                         toAdd);
    }
  }
  if (ind2 > 0) {
    j = B[ind2] - '0';
    for (i = 0; i < 10; i++) {
      k = (i + j + carry) % 10;
      int toAdd = 0, nind1 = ind1, nind2 = ind2, nind3 = ind3,
          ncarry = (i + j + carry > 9);
      if (ind1 > 0 && i == A[ind1] - '0')
        nind1--;
      else
        toAdd++;
      if (ind2 > 0 && j == B[ind2] - '0')
        nind2--;
      else
        toAdd++;
      if (ind3 > 0 && k == C[ind3] - '0')
        nind3--;
      else
        toAdd++;
      ret = min(ret, solve(nind1, nind2, nind3, (i ? 0 : (l1 + 1)),
                           (j ? 0 : (l2 + 1)), (k ? 0 : (l3 + 1)), ncarry) +
                         toAdd);
    }
  }
  if (ind3 > 0) {
    k = C[ind3] - '0';
    for (i = 0; i < 10; i++) {
      j = (k - i - carry + 10) % 10;
      int toAdd = 0, nind1 = ind1, nind2 = ind2, nind3 = ind3,
          ncarry = (i + j + carry > 9);
      if (ind1 > 0 && i == A[ind1] - '0')
        nind1--;
      else
        toAdd++;
      if (ind2 > 0 && j == B[ind2] - '0')
        nind2--;
      else
        toAdd++;
      if (ind3 > 0 && k == C[ind3] - '0')
        nind3--;
      else
        toAdd++;
      ret = min(ret, solve(nind1, nind2, nind3, (i ? 0 : (l1 + 1)),
                           (j ? 0 : (l2 + 1)), (k ? 0 : (l3 + 1)), ncarry) +
                         toAdd);
    }
  }
  return best[ind1][ind2][ind3][l1][l2][l3][carry] = ret;
}
void solve2(int ind1, int ind2, int ind3, int l1, int l2, int l3, bool carry) {
  if (ind1 == 0 && ind2 == 0 && ind3 == 0) {
    if (carry) ret3 += '1';
    return;
  }
  int i, j, k, cur = best[ind1][ind2][ind3][l1][l2][l3][carry];
  if (ind1 > 0) {
    i = A[ind1] - '0';
    for (j = 0; j < 10; j++) {
      k = (i + j + carry) % 10;
      int toAdd = 0, nind1 = ind1, nind2 = ind2, nind3 = ind3,
          ncarry = (i + j + carry > 9);
      if (ind1 > 0 && i == A[ind1] - '0')
        nind1--;
      else
        toAdd++;
      if (ind2 > 0 && j == B[ind2] - '0')
        nind2--;
      else
        toAdd++;
      if (ind3 > 0 && k == C[ind3] - '0')
        nind3--;
      else
        toAdd++;
      if (solve(nind1, nind2, nind3, i ? 0 : (l1 + 1), j ? 0 : (l2 + 1),
                k ? 0 : (l3 + 1), ncarry) +
              toAdd ==
          cur) {
        ret1 += char(i + '0');
        ret2 += char(j + '0');
        ret3 += char(k + '0');
        solve2(nind1, nind2, nind3, (i ? 0 : (l1 + 1)), (j ? 0 : (l2 + 1)),
               (k ? 0 : (l3 + 1)), ncarry);
        return;
      }
    }
  }
  if (ind2 > 0) {
    j = B[ind2] - '0';
    for (i = 0; i < 10; i++) {
      k = (i + j + carry) % 10;
      int toAdd = 0, nind1 = ind1, nind2 = ind2, nind3 = ind3,
          ncarry = (i + j + carry > 9);
      if (ind1 > 0 && i == A[ind1] - '0')
        nind1--;
      else
        toAdd++;
      if (ind2 > 0 && j == B[ind2] - '0')
        nind2--;
      else
        toAdd++;
      if (ind3 > 0 && k == C[ind3] - '0')
        nind3--;
      else
        toAdd++;
      if (solve(nind1, nind2, nind3, i ? 0 : (l1 + 1), j ? 0 : (l2 + 1),
                k ? 0 : (l3 + 1), ncarry) +
              toAdd ==
          cur) {
        ret1 += char(i + '0');
        ret2 += char(j + '0');
        ret3 += char(k + '0');
        solve2(nind1, nind2, nind3, (i ? 0 : (l1 + 1)), (j ? 0 : (l2 + 1)),
               (k ? 0 : (l3 + 1)), ncarry);
        return;
      }
    }
  }
  if (ind3 > 0) {
    k = C[ind3] - '0';
    for (i = 0; i < 10; i++) {
      j = (k - i - carry + 10) % 10;
      int toAdd = 0, nind1 = ind1, nind2 = ind2, nind3 = ind3,
          ncarry = (i + j + carry > 9);
      if (ind1 > 0 && i == A[ind1] - '0')
        nind1--;
      else
        toAdd++;
      if (ind2 > 0 && j == B[ind2] - '0')
        nind2--;
      else
        toAdd++;
      if (ind3 > 0 && k == C[ind3] - '0')
        nind3--;
      else
        toAdd++;
      if (solve(nind1, nind2, nind3, i ? 0 : (l1 + 1), j ? 0 : (l2 + 1),
                k ? 0 : (l3 + 1), ncarry) +
              toAdd ==
          cur) {
        ret1 += char(i + '0');
        ret2 += char(j + '0');
        ret3 += char(k + '0');
        solve2(nind1, nind2, nind3, (i ? 0 : (l1 + 1)), (j ? 0 : (l2 + 1)),
               (k ? 0 : (l3 + 1)), ncarry);
        return;
      }
    }
  }
}
string removeLeadingZeros(string x) {
  for (int i = 0; i < (int)x.size(); i++)
    if (x[i] != '0') return x.substr(i);
  return x;
}
int main() {
  memset(best, -1, sizeof best);
  scanf("%d+%d=%d", &a, &b, &c);
  sprintf(A, "#%d", a);
  sprintf(B, "#%d", b);
  sprintf(C, "#%d", c);
  solve(strlen(A) - 1, strlen(B) - 1, strlen(C) - 1, 0, 0, 0, 0);
  solve2(strlen(A) - 1, strlen(B) - 1, strlen(C) - 1, 0, 0, 0, 0);
  reverse(ret1.begin(), ret1.end());
  reverse(ret2.begin(), ret2.end());
  reverse(ret3.begin(), ret3.end());
  ret1 = removeLeadingZeros(ret1);
  ret2 = removeLeadingZeros(ret2);
  ret3 = removeLeadingZeros(ret3);
  printf("%s+%s=%s\n", ret1.c_str(), ret2.c_str(), ret3.c_str());
}