Problem:
Score : 300 points

Problem Statement

You are given a string S consisting of digits between 1 and 9, inclusive.
You can insert the letter + into some of the positions (possibly none) between two letters in this string.
Here, + must not occur consecutively after insertion.

All strings that can be obtained in this way can be evaluated as formulas.

Evaluate all possible formulas, and print the sum of the results.

Constraints

1 \leq |S| \leq 10

All letters in S are digits between 1 and 9, inclusive.

Input

The input is given from Standard Input in the following format:

S

Output

Print the sum of the evaluated value over all possible formulas.

Sample Input 1

125

Sample Output 1

176

There are 4 formulas that can be obtained: 125, 1+25, 12+5 and 1+2+5. When each formula is evaluated,

125

1+25=26

12+5=17

1+2+5=8

Thus, the sum is 125+26+17+8=176.

Sample Input 2

9999999999

Sample Output 2

12656242944

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<sstream>
#include<string>
#include<numeric>
#include <algorithm>
#include<math.h>
#include<cstdio>
#include<string.h>
#include<unistd.h>
#include <array>
#include <map> 




using namespace std;
typedef long long ll;
struct point{
  int x;
  int y;

};
int gcd(int a, int b)
{
	return a == 0 ? b : gcd(b % a, a);
}
int lcm( int m, int n )
{
	// 引数に０がある場合は０を返す
	if ( ( 0 == m ) || ( 0 == n ) )
		return 0;
	
	return ((m / gcd(m, n)) * n); // lcm = m * n / gcd(m,n)
}//lcm
int input(){
  int x;
  cin>>x;
  return x;
}

int moji(char in)
{
    int ans = (int)in-(int)'a';
    if((ans < 0) || (ans > 25)){
        ans = 26;
    }
    return ans;
}
const int VV=10;//場合に応じてVVの値のみ変更する必要あり
//dijkstra(s)sがスタート地点でそこからの最短距離を配列dで表す。正の重みのみ使用可能
int cost[VV][VV];
int d[VV];
bool used[VV];
void dijkstra(int s){
  fill(d,d+VV,100000);
  fill(used,used+VV,false);
  d[s]=0;
  while(true){
    int v=-1;
    for(int u=0;u<VV;u++){
      if(!used[u]&&(v==-1||d[u]<d[v]))v=u;
    }
    if(v==-1)break;
    used[v]=true;
    for(int u=0;u<VV;u++){
      d[u]=min(d[u],d[v]+cost[v][u]);
    }
  }

}

int compare_int(const void *a, const void *b)//qsort(quick sort利用時に使用)
{
    return *(int*)a - *(int*)b;
}

int binary_searchh(long long x,long long k[],int n){
  int l=0;
  int r=n;
  while(r-l>=1){
    int i=(l+r)/2;
    if(k[i]==x)return i;
    else if(k[i]<x)l=i+1;
    else r=i;
  }
  return -1;
}

struct File {
    int aa;
    int bb;

    File(const int& aa, const int& bb)
        : aa(aa), bb(bb) {}
};

bool operator<(const File& a, const File& b)
{
    // ファイル種別、ファイル名の順番で優先順位を付けて比較
    return std::tie(a.aa, a.bb) < std::tie(b.aa, b.bb);
}

long long kaijo(long long x){
  long long l=10*10*10*10*10*10*10*10*10+7;
  long long sum=1;
  for(int i=x;i>0;i--){
    sum*=i;
    if(sum>l){
      sum%=l;
    }
  }
  return sum;
}






int main(){
  string S;
  cin>>S;
  long long N=S.size();
  long long sum=0;
  for(int i=0;i<1<<N;i++){
    long long tmp=0;
    for(int j=0;j<N;j++){
      if(i&1<<j){
        sum+=tmp;
        tmp=0;

      }else{
        tmp=tmp*10+S[j]-'0';
      }
    }
    sum+=tmp;
  }
  cout<<sum<<endl;

}