Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya loves long lucky numbers very much. He is interested in the minimum lucky number d that meets some condition. Let cnt(x) be the number of occurrences of number x in number d as a substring. For example, if d = 747747, then cnt(4) = 2, cnt(7) = 4, cnt(47) = 2, cnt(74) = 2. Petya wants the following condition to fulfil simultaneously: cnt(4) = a1, cnt(7) = a2, cnt(47) = a3, cnt(74) = a4. Petya is not interested in the occurrences of other numbers. Help him cope with this task.

Input

The single line contains four integers a1, a2, a3 and a4 (1 ≤ a1, a2, a3, a4 ≤ 106).

Output

On the single line print without leading zeroes the answer to the problem — the minimum lucky number d such, that cnt(4) = a1, cnt(7) = a2, cnt(47) = a3, cnt(74) = a4. If such number does not exist, print the single number "-1" (without the quotes).

Examples

Input

2 2 1 1


Output

4774


Input

4 7 3 1


Output

-1
#include <bits/stdc++.h>
using namespace std;
deque<int> dq;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j, k, l, n, m, p, q, first = -1, la = -1;
  cin >> n >> m >> p >> q;
  int flag = 0;
  if (abs(p - q) > 1)
    flag = 1;
  else {
    if (p > q) {
      for (i = 0; i < p; i++) {
        dq.push_back(4);
        dq.push_back(7);
        n--;
        m--;
      }
      if (n < 0 || m < 0)
        flag = 1;
      else {
        int sz = dq.size();
        for (i = 0; i < sz; i++) {
          if (dq[i] == 4 && first == -1)
            first = i;
          else if (dq[i] == 7)
            la = i;
        }
        for (i = 0; i < sz; i++) {
          cout << dq[i];
          if (i == first) {
            while (n--) cout << '4';
          }
          if (i == la) {
            while (m--) cout << '7';
          }
        }
      }
    } else if (q > p) {
      for (i = 0; i < q; i++) {
        dq.push_back(7);
        dq.push_back(4);
        n--;
        m--;
      }
      if (n < 0 || m < 0)
        flag = 1;
      else {
        int sz = dq.size();
        for (i = 0; i < sz; i++) {
          if (dq[i] == 4 && first == -1)
            first = i;
          else if (dq[i] == 7)
            la = i;
        }
        for (i = 0; i < sz; i++) {
          cout << dq[i];
          if (i == first) {
            while (n--) cout << '4';
          }
          if (i == la) {
            while (m--) cout << '7';
          }
        }
      }
    } else {
      if (p == 1) {
        if (n > 1) {
          dq.push_back(4);
          dq.push_back(7);
          dq.push_back(4);
          n -= 2;
          m--;
        } else {
          dq.push_back(7);
          dq.push_back(4);
          dq.push_back(7);
          n -= 1;
          m -= 2;
        }
      } else {
        if (n >= p + 1) {
          for (i = 0; i < p; i++) {
            dq.push_back(4);
            dq.push_back(7);
            n--;
            m--;
          }
          dq.push_back(4);
          n--;
        } else {
          for (i = 0; i < p; i++) {
            dq.push_back(7);
            dq.push_back(4);
            n--;
            m--;
          }
          dq.push_back(7);
          m--;
        }
      }
      if (n < 0 || m < 0)
        flag = 1;
      else {
        int sz = dq.size();
        for (i = 0; i < sz; i++) {
          if (dq[i] == 4 && first == -1)
            first = i;
          else if (dq[i] == 7)
            la = i;
        }
        for (i = 0; i < sz; i++) {
          cout << dq[i];
          if (i == first) {
            while (n--) cout << '4';
          }
          if (i == la) {
            while (m--) cout << '7';
          }
        }
      }
    }
  }
  if (flag == 1) cout << "-1";
  return 0;
}
