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;
template <class T>
inline T modinv(T a, T n) {
  T i = n, v = 0, d = 1;
  while (a > 0) {
    T t = i / a, x = a;
    a = i % x;
    i = x;
    x = d;
    d = v - t * x;
    v = x;
  }
  return (v + n) % n;
}
long long modpow(long long n, long long k, long long mod) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % mod;
    k >>= 1;
    n = (n * n) % mod;
  }
  return ans % mod;
}
template <class T>
struct Fraction {
  T a, b;
  Fraction(T a = 0, T b = 1);
  string toString();
};
template <class T>
Fraction<T>::Fraction(T a, T b) {
  T d = GCD(a, b);
  a /= d;
  b /= d;
  if (b < 0) a = -a, b = -b;
  this->a = a;
  this->b = b;
}
template <class T>
string Fraction<T>::toString() {
  ostringstream sout;
  sout << a << "/" << b;
  return sout.str();
}
template <class T>
Fraction<T> operator+(Fraction<T> p, Fraction<T> q) {
  return Fraction<T>(p.a * q.b + q.a * p.b, p.b * q.b);
}
template <class T>
Fraction<T> operator-(Fraction<T> p, Fraction<T> q) {
  return Fraction<T>(p.a * q.b - q.a * p.b, p.b * q.b);
}
template <class T>
Fraction<T> operator*(Fraction<T> p, Fraction<T> q) {
  return Fraction<T>(p.a * q.a, p.b * q.b);
}
template <class T>
Fraction<T> operator/(Fraction<T> p, Fraction<T> q) {
  return Fraction<T>(p.a * q.b, p.b * q.a);
}
template <class T>
string str(T Number) {
  string Result;
  ostringstream convert;
  convert << Number;
  Result = convert.str();
  return Result;
}
int StringToNumber(const string &Text) {
  istringstream ss(Text);
  int result;
  return ss >> result ? result : 0;
}
template <class T>
inline vector<pair<T, int> > FACTORISE(T n) {
  vector<pair<T, int> > R;
  for (T i = 2; n > 1;) {
    if (n % i == 0) {
      int C = 0;
      for (; n % i == 0; C++, n /= i)
        ;
      R.push_back(make_pair(i, C));
    }
    i++;
    if (i > n / i) i = n;
  }
  if (n > 1) R.push_back(make_pair(n, 1));
  return R;
}
template <class T>
inline T TOTIENT(T n) {
  vector<pair<T, int> > R = FACTORISE(n);
  T r = n;
  for (int i = 0; i < R.size(); i++) r = r / R[i].first * (R[i].first - 1);
  return r;
}
template <class T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
double rnd(float d) { return floor(d + 0.49); }
char s[100005], ans[100005];
int n;
void go(int m) {
  for (int i = 0; i < m / 2; i++) ans[i] = '4';
  for (int i = m / 2; i < m; i++) ans[i] = '7';
  ans[m] = 0;
}
bool solve(int i, int c4, bool greater) {
  if (i == n) return c4 == 0;
  if (greater) {
    if (c4) {
      ans[i] = '4';
      return solve(i + 1, c4 - 1, true);
    } else {
      ans[i] = '7';
      return solve(i + 1, c4, true);
    }
  } else {
    if (s[i] < '4') {
      if (c4) {
        ans[i] = '4';
        return solve(i + 1, c4 - 1, true);
      } else {
        ans[i] = '7';
        return solve(i + 1, c4, true);
      }
    } else if (s[i] == '4') {
      if (c4) {
        ans[i] = '4';
        if (solve(i + 1, c4 - 1, false)) return true;
      }
      ans[i] = '7';
      return solve(i + 1, c4, true);
    } else if (s[i] <= '7') {
      ans[i] = '7';
      return solve(i + 1, c4, s[i] < '7');
    }
    return false;
  }
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  if (n % 2 == 0) {
    if (!solve(0, n / 2, false))
      go(n + 2);
    else
      ans[n] = 0;
  } else {
    go(n + 1);
  }
  printf("%s\n", ans);
  return 0;
}