Problem: Petya loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.

One day Petya came across a positive integer n. Help him to find the least super lucky number which is not less than n.

Input

The only line contains a positive integer n (1 ≤ n ≤ 10100000). This number doesn't have leading zeroes.

Output

Output the least super lucky number that is more than or equal to n.

Examples

Input

4500


Output

4747


Input

47


Output

47

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int INF = 2e18;
const long long int N = 1e6;
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
long long int fact(long long int n) {
  long long int ans = 1;
  for (long long int i = 1; i <= n; i++) ans = (ans * i) % 1000000000;
  return ans;
}
long long int exp(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2 == 1) res = (res * x) % 1000000000;
    x = (x * x) % 1000000000;
    n = n / 2;
  }
  return res;
}
void null() {
  long long int n;
  cin >> n;
  string s = to_string(n);
  long long int m = s.size() % 2 ? s.size() + 1 : s.size();
  vector<long long int> v;
  string z = "";
  for (int i = 0; i < m / 2; i++) z += '4';
  for (int i = 0; i < m / 2; i++) z += '7';
  do {
    long long int ans = stoll(z);
    v.push_back(ans);
  } while (next_permutation(z.begin(), z.end()));
  z += '4';
  z += '7';
  sort(z.begin(), z.end());
  v.push_back(stoll(z));
  cout << (*lower_bound(v.begin(), v.end(), n)) << '\n';
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(9);
  cout << fixed;
  long long int t = 1;
  clock_t start, end;
  start = clock();
  while (t--) {
    null();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}