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 int MOD = 1e9 + 7;
const long long LINF = 2e16;
const int INF = 2e9;
const int magic = 348;
const double eps = 1e-10;
const double pi = 3.14159265;
inline int getint() {
  char ch;
  int res;
  bool f;
  while (!isdigit(ch = getchar()) && ch != '-') {
  }
  if (ch == '-')
    f = false, res = 0;
  else
    f = true, res = ch - '0';
  while (isdigit(ch = getchar())) res = res * 10 + ch - '0';
  return f ? res : -res;
}
struct STATE {
  int p1, p2, p3, cap, e1, e2, e3;
  inline void print() {
    cout << p1 << ' ' << p2 << ' ' << p3 << ' ' << cap << ' ' << e1 << ' ' << e2
         << ' ' << e3 << endl;
  }
};
inline int ToState(int p1, int p2, int p3, int cap, int e1, int e2, int e3) {
  return (e1 * 4 + e2 * 2 + e3) * 2000 + cap * 1000 + p1 * 100 + p2 * 10 + p3;
}
inline STATE ToNUM(int cur) {
  STATE res = {0, 0, 0, 0, 0, 0, 0};
  int pt1 = cur / 2000, pt2 = cur % 2000;
  res.cap = pt2 / 1000;
  res.p1 = pt2 / 100 % 10;
  res.p2 = pt2 / 10 % 10;
  res.p3 = pt2 % 10;
  res.e1 = pt1 / 4;
  res.e2 = pt1 / 2 % 2;
  res.e3 = pt1 % 2;
  return res;
}
int dist[20048], num[4][20], len[4];
pair<int, int> from[20048];
string s;
int head[500048], nxt[1000048], to[1000048], f[1000048], method[1000048],
    tot = 1;
inline void add_edge(int s, int t, int c, int way) {
  to[++tot] = t;
  nxt[tot] = head[s];
  head[s] = tot;
  f[tot] = c;
  method[tot] = way;
}
inline void STRING_TO_NUM(string s, int ind) {
  int i, pt = 0;
  for (i = int(s.size()) - 1; i >= 0; i--) num[ind][++pt] = s[i] - '0';
  len[ind] = pt;
}
inline pair<bool, int> check(int p1, int p2, int p3, int cap, int add1,
                             int add2, int add3, int tocap, int e1, int e2,
                             int e3, int le1, int le2, int le3) {
  int i, j, k, n1, n2, n3;
  for ((e1 && !le1) ? i = 1 : i = 0; le1 ? i <= 0 : i <= 9; i++)
    for ((e2 && !le2) ? j = 1 : j = 0; le2 ? j <= 0 : j <= 9; j++)
      for ((e3 && !le3) ? k = 1 : k = 0; le3 ? k <= 0 : k <= 9; k++) {
        if (tocap && e3) continue;
        n1 = (add1 ? num[1][p1 + 1] : i);
        n2 = (add2 ? num[2][p2 + 1] : j);
        n3 = (add3 ? num[3][p3 + 1] : k);
        if (n1 + n2 + cap == tocap * 10 + n3)
          return make_pair(true, n1 * 100 + n2 * 10 + n3);
      }
  return make_pair(false, -1);
}
inline int count_num(int e1, int e2, int e3) { return 3 - e1 - e2 - e3; }
void addedge(int p1, int p2, int p3, int cap, int e1, int e2, int e3) {
  if ((p1 < len[1] && e1) || (p2 < len[2] && e2) || (p3 < len[3] && e3)) return;
  int cur = ToState(p1, p2, p3, cap, e1, e2, e3), i, j, k, tocap, toe1, toe2,
      toe3;
  for (i = 0; i <= 1; i++)
    for (j = 0; j <= 1; j++)
      for (k = 0; k <= 1; k++)
        for (tocap = 0; tocap <= 1; tocap++)
          for (toe1 = 0; toe1 <= 1; toe1++)
            for (toe2 = 0; toe2 <= 1; toe2++)
              for (toe3 = 0; toe3 <= 1; toe3++) {
                if ((p1 == len[1] && i) || (p2 == len[2] && j) ||
                    (p3 == len[3] && k))
                  continue;
                if ((p1 + i < len[1] && toe1) || (p2 + j < len[2] && toe2) ||
                    (p3 + k < len[3] && toe3))
                  continue;
                if ((e1 && !toe1) || (e2 && !toe2) || (e3 && !toe3)) continue;
                pair<bool, int> res = check(p1, p2, p3, cap, i, j, k, tocap,
                                            toe1, toe2, toe3, e1, e2, e3);
                if (!res.first) continue;
                add_edge(
                    cur,
                    ToState(p1 + i, p2 + j, p3 + k, tocap, toe1, toe2, toe3),
                    count_num(e1, e2, e3), res.second);
              }
}
priority_queue<pair<int, int> > q;
void dijkstra() {
  int i, second, dd, cur, starter = ToState(0, 0, 0, 0, 0, 0, 0);
  q.push(make_pair(0, starter));
  for (i = 0; i <= 20000; i++) dist[i] = INF;
  dist[starter] = 0;
  from[starter] = make_pair(-1, -1);
  while (!q.empty()) {
    dd = -q.top().first;
    cur = q.top().second;
    q.pop();
    if (dd > dist[cur]) continue;
    for (i = head[cur]; i; i = nxt[i]) {
      second = to[i];
      if (dist[cur] + f[i] < dist[second]) {
        dist[second] = dist[cur] + f[i];
        from[second] = make_pair(cur, method[i]);
        q.push(make_pair(-dist[second], second));
      }
    }
  }
}
int ans1, ans2, ans3;
void construct_ans(int ed) {
  ans1 = ans2 = ans3 = 0;
  int cur = ed;
  while (cur != -1) {
    if (from[cur].first == -1) break;
    ans1 = ans1 * 10 + from[cur].second / 100;
    ans2 = ans2 * 10 + from[cur].second / 10 % 10;
    ans3 = ans3 * 10 + from[cur].second % 10;
    cur = from[cur].first;
  }
}
int main() {
  cin >> s;
  int pos = s.find('+'), pos1;
  STRING_TO_NUM(s.substr(0, pos), 1);
  pos1 = pos + 1;
  pos = s.find('=', pos1);
  STRING_TO_NUM(s.substr(pos1, pos - pos1), 2);
  STRING_TO_NUM(s.substr(pos + 1, int(s.size()) - 1 - (pos + 1) + 1), 3);
  int i, j, k, cap, e1, e2, e3;
  for (i = 0; i <= len[1]; i++)
    for (j = 0; j <= len[2]; j++)
      for (k = 0; k <= len[3]; k++)
        for (cap = 0; cap <= 1; cap++)
          for (e1 = 0; e1 <= 1; e1++)
            for (e2 = 0; e2 <= 1; e2++)
              for (e3 = 0; e3 <= 1; e3++) addedge(i, j, k, cap, e1, e2, e3);
  dijkstra();
  int ed;
  if (dist[ToState(len[1], len[2], len[3], 0, 1, 1, 1)] <
      dist[ToState(len[1], len[2], len[3], 1, 1, 1, 1)])
    ed = ToState(len[1], len[2], len[3], 0, 1, 1, 1);
  else
    ed = ToState(len[1], len[2], len[3], 1, 1, 1, 1);
  construct_ans(ed);
  printf("%d+%d=%d\n", ans1, ans2, ans3);
  return 0;
}