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;
int a1, a2, a3, a4, p, mn;
const int N = 1e5 + 10, mod = 1e9 + 7;
string s, s4, s7, ans;
bool ff = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int r = 0;
  cin >> a1 >> a2 >> a3 >> a4;
  if (abs(a3 - a4) > 1) return cout << -1, 0;
  mn = min(a3, a4);
  while (mn--) {
    s += "47";
    a1--;
    a2--;
  }
  mn = min(a3, a4);
  a3 -= mn;
  a4 -= mn;
  if (a1) {
    s += '4';
    a1--;
  } else if (a2) {
    reverse(s.begin(), s.end());
    s += '7';
    a2--;
  } else
    ff = 1;
  if (a4) {
    if (s[s.size() - 1] == '7')
      s += '4', a1--;
    else
      s = "7" + s, a2--;
    a4--;
  } else if (a3) {
    if (s[s.size() - 1] == '4')
      s += '7', a2--;
    else
      ff = 1;
    a3--;
  }
  for (int i = 1; i <= a1; i++) s4 += '4';
  int j = s.size() - 1, i = 0, k = 0, p = 0;
  for (int i = 1; i <= a2; i++) s7 += '7';
  if (!(a1 >= 0 && a2 >= 0) || ff) return cout << -1, 0;
  for (; j >= 0; j--)
    if (s[j] == '7') {
      p = j;
      break;
    }
  for (; i < s.size(); i++)
    if (s[i] == '4') {
      k = i;
      break;
    }
  for (int r = 0; r < s.size(); r++) {
    if (r == p) cout << s7;
    if (r == k) cout << s4;
    cout << s[r];
  }
}
