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 MAX = 8;
const int inf = 2e9;
int sel[MAX][MAX][MAX][2][2][2];
struct edge {
  int nA, nB, nC;
  int f1, f2, c;
  int cost;
  int sela, selb;
  edge() {}
  edge(int a, int b, int c, int d, int e, int f, int g, int h, int i)
      : nA(a), nB(b), nC(c), f1(d), f2(e), c(f), cost(g), sela(h), selb(i) {}
  void print() {
    printf("%d %d %d %d %d %d c = %d %d %d\n", nA, nB, nC, f1, f2, c, cost,
           sela, selb);
  }
};
vector<edge> adj[MAX][MAX][MAX][2][2][2];
int a, b, c;
int A[MAX], B[MAX], C[MAX];
int Alen, Blen, Clen;
int mA, mB, mC;
int dp[MAX][MAX][MAX][2][2][2];
inline void subadd(int tA, int tB, int tC, int tf1, int tf2, int tc, int nA,
                   int nB, int nC, int f1, int f2, int c, int cost, int sela,
                   int selb) {
  if (tA != nA || tB != nB || tC != nC) {
    adj[tA][tB][tC][tf1][tf2][tc].push_back(
        {nA, nB, nC, f1, f2, c, cost, sela, selb});
  }
}
inline void add(int nA, int nB, int nC, int f1, int f2, int c, int adda,
                int addb) {
  int t1, t2, cc, cost;
  if (nA == Alen)
    mA = Alen;
  else {
    if (adda == -1)
      mA = nA;
    else if (adda == A[nA])
      mA = nA + 1;
    else
      mA = nA;
  }
  if (nB == Blen)
    mB = Blen;
  else {
    if (addb == -1)
      mB = nB;
    else if (addb == B[nB])
      mB = nB + 1;
    else
      mB = nB;
  }
  t1 = (adda == -1 ? 0 : adda) + (addb == -1 ? 0 : addb) + c;
  t2 = t1 % 10;
  if (nC == Clen)
    mC = Clen;
  else if (t2 == C[nC])
    mC = nC + 1;
  else
    mC = nC;
  cost = (int)(adda != -1) + (int)(addb != -1);
  if (t1 >= 10) {
    cc = 1;
    if (c == 1)
      cost++;
    else
      cost += 2;
  } else {
    cc = 0;
    if (c != 1) cost++;
  }
  if (f1) {
    if (addb == 0) {
      subadd(mA, mB, mC, f1, 0, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
    } else {
      subadd(mA, mB, mC, f1, 0, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
      subadd(mA, mB, mC, f1, 1, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
    }
  } else if (f2) {
    if (adda == 0) {
      subadd(mA, mB, mC, 0, f2, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
    } else {
      subadd(mA, mB, mC, 0, f2, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
      subadd(mA, mB, mC, 1, f2, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
    }
  } else {
    if (adda == 0 && addb == 0) {
      subadd(mA, mB, mC, 0, 0, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
    } else if (adda == 0 && addb) {
      subadd(mA, mB, mC, 0, 1, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
      subadd(mA, mB, mC, 0, 0, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
    } else if (adda && addb == 0) {
      subadd(mA, mB, mC, 0, 0, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
      subadd(mA, mB, mC, 1, 0, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
    } else {
      subadd(mA, mB, mC, 0, 0, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
      subadd(mA, mB, mC, 0, 1, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
      subadd(mA, mB, mC, 1, 0, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
      subadd(mA, mB, mC, 1, 1, cc, nA, nB, nC, f1, f2, c, cost, adda, addb);
    }
  }
}
void init() {
  int temp;
  temp = a;
  Alen = Blen = Clen = 0;
  while (a) {
    A[Alen++] = a % 10;
    a /= 10;
  }
  temp = b;
  while (b) {
    B[Blen++] = b % 10;
    b /= 10;
  }
  temp = c;
  while (c) {
    C[Clen++] = c % 10;
    c /= 10;
  }
  for (int nA = (0); nA <= (Alen); ++nA) {
    for (int nB = (0); nB <= (Blen); ++nB) {
      for (int nC = (0); nC <= (Clen); ++nC) {
        for (int f1 = (0); f1 <= (1); ++f1) {
          for (int f2 = (0); f2 <= (1); ++f2) {
            for (int c = (0); c <= (1); ++c) {
              if (f1 == 0 || f2 == 0) {
                if (f1 == 0 && f2 == 0) {
                  for (int adda = (0); adda <= (9); ++adda) {
                    for (int addb = (0); addb <= (9); ++addb) {
                      add(nA, nB, nC, f1, f2, c, adda, addb);
                    }
                  }
                } else if (f1 == 0) {
                  for (int adda = (0); adda <= (9); ++adda) {
                    add(nA, nB, nC, f1, f2, c, adda, -1);
                  }
                } else {
                  for (int addb = (0); addb <= (9); ++addb) {
                    add(nA, nB, nC, f1, f2, c, -1, addb);
                  }
                }
              }
            }
          }
        }
      }
    }
  }
}
int dfs(int tA, int tB, int tC, int f1, int f2, int c) {
  if (dp[tA][tB][tC][f1][f2][c] != inf) return dp[tA][tB][tC][f1][f2][c];
  int ans = inf, t1, t2;
  for (int i = 0; i < adj[tA][tB][tC][f1][f2][c].size(); i++) {
    edge &e = adj[tA][tB][tC][f1][f2][c][i];
    if (!(tA == e.nA && tB == e.nB && tC == e.nC && f1 == e.f1 && f2 == e.f2 &&
          c == e.c)) {
      t1 = dfs(e.nA, e.nB, e.nC, e.f1, e.f2, e.c);
      if (t1 == inf) continue;
      t2 = t1 + e.cost;
      if (t2 < ans) {
        ans = t2;
        sel[tA][tB][tC][f1][f2][c] = i;
      }
    }
  }
  return dp[tA][tB][tC][f1][f2][c] = ans;
}
int main() {
  scanf("%d + %d = %d\n", &a, &b, &c);
  memset(sel, -1, sizeof(sel));
  init();
  for (int nA = (0); nA <= (Alen); ++nA)
    for (int nB = (0); nB <= (Blen); ++nB)
      for (int nC = (0); nC <= (Clen); ++nC)
        for (int f1 = (0); f1 <= (1); ++f1)
          for (int f2 = (0); f2 <= (1); ++f2)
            for (int c = (0); c <= (1); ++c) dp[nA][nB][nC][f1][f2][c] = inf;
  dp[0][0][0][0][0][0] = 0;
  int sA = Alen, sB = Blen, sC = Clen, f1 = 1, f2 = 1, c;
  int t1 = dfs(Alen, Blen, Clen, 1, 1, 0);
  int t2 = dfs(Alen, Blen, Clen, 1, 1, 1);
  int t3;
  if (C[Clen - 1] == 1)
    t3 = dfs(Alen, Blen, Clen - 1, 1, 1, 1);
  else
    t3 = inf;
  if (t1 < t2) {
    c = 0;
    if (t3 < t1) --sC, c = 1;
  } else {
    c = 1;
    if (t3 < t2) --sC;
  }
  vector<int> ansA, ansB;
  assert(sel[sA][sB][sC][f1][f2][c] != -1);
  while (sel[sA][sB][sC][f1][f2][c] != -1) {
    int i = sel[sA][sB][sC][f1][f2][c];
    edge &e = adj[sA][sB][sC][f1][f2][c][i];
    if (e.sela != -1) ansA.push_back(e.sela);
    if (e.selb != -1) ansB.push_back(e.selb);
    sA = e.nA, sB = e.nB, sC = e.nC, f1 = e.f1, f2 = e.f2, c = e.c;
  }
  int A = 0, B = 0, C;
  for (int i = (0); i <= ((int)ansA.size() - 1); ++i) A = A * 10 + ansA[i];
  for (int i = (0); i <= ((int)ansB.size() - 1); ++i) B = B * 10 + ansB[i];
  C = A + B;
  printf("%d+%d=%d\n", A, B, C);
  return 0;
}