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

Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.

Input

The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.

Output

In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).

Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

2 7


Output

33


Input

7 7


Output

7

Note

In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33

In the second sample: next(7) = 7
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int INF = (int)1e9;
const int EPS = 1e-6;
const int mod = 1000000007;
const long double PI = 3.14159265359;
const int dx4[4] = {0, 1, 0, -1};
const int dy4[4] = {-1, 0, 1, 0};
const int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
inline bool grid_inside(int x, int y, int n, int m) {
  return x >= 1 && x <= n && y >= 1 && y <= m;
}
template <typename T>
T lcm(T a, T b) {
  return (a * b) / __gcd(a, b);
}
template <typename T>
T Dis(T x1, T y1, T x2, T y2) {
  return sqrt(((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2)));
}
template <typename T>
T Angle(T x1, T y1, T x2, T y2) {
  return atan((double)(y1 - y2) / (double)(x1 - x2));
}
inline int Set(int N, int pos) { return N = N | (1 << pos); }
inline int Reset(int N, int pos) { return N = N & ~(1 << pos); }
inline bool Check(int N, int pos) { return (bool)(N & (1 << pos)); }
template <class T, class X>
inline T togglebit(T a, X i) {
  T t = 1;
  return (a ^ (t << i));
}
bool compare(const pair<int, int>& i, const pair<int, int>& j) {
  return i.first < j.first;
}
int LSOne(long long int n) { return log2(n & -n) + 1; }
int rangeClear(int n, int i, int j) {
  int ones = ~0;
  int a = ones << (j + 1);
  int b = (i << 2) - 1;
  int mask = a | b;
  int ans = n & mask;
  return ans;
}
long long int mul_mod(long long int x, long long int y, long long int mod) {
  long long int i = 0, j = x;
  while (y > 0) {
    if (y & 1) {
      i = (i + j) % mod;
    }
    j = (j + j) % mod;
    y >>= 1;
  }
  return i;
}
long long int fast_power(long long int a, int x) {
  if (x == 0)
    return 1;
  else if (x == 1)
    return a;
  else {
    long long int R = fast_power(a, x >> 1);
    if (!(x & 1))
      return R * R;
    else
      return R * a * R;
  }
}
long long int fast_power_unlocked(long long int a, long long int n,
                                  long long int mod) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) {
      res = mul_mod(res, a, mod);
    }
    a = mul_mod(a, a, mod);
    n = n >> 1;
  }
  return res;
}
long long int modInverse(long long int A, long long int M) {
  return fast_power_unlocked(A, M - 2, M);
}
const int maxn = 1000;
void solve() {
  long long int l, r;
  cin >> l >> r;
  vector<long long int> v;
  long long int ans = 0;
  for (int i = 1; i <= 9; i++) {
    int mask = (1 << i);
    for (int j = 0; j < mask; j++) {
      bitset<9> b(j);
      string s = "";
      if (j == 0) {
        for (int k = 0; k < i; k++) {
          s += "4";
        }
      } else {
        for (int k = i - 1; k >= 0; k--) {
          s += (b[k]) ? "7" : "4";
        }
      }
      v.emplace_back(stoll(s));
    }
  }
  v.emplace_back(4444444444);
  int sz = v.size();
  for (int i = 0; i < sz; i++) {
    if (v[i] >= l) {
      if (v[i] >= r) {
        ans += ((r - l + 1) * v[i]);
        break;
      }
      ans += (v[i] - l + 1) * v[i];
      l = v[i] + 1;
    }
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  for (auto blockTime = make_pair(chrono::high_resolution_clock::now(), true);
       blockTime.second;
       fprintf(stderr, "%s: %ld ms\n", "Time Taken",
               chrono::duration_cast<chrono::milliseconds>(
                   chrono::high_resolution_clock::now() - blockTime.first)
                   .count()),
            fflush(stderr), blockTime.second = false) {
    solve();
  }
}
