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 (GCC 9.2.1) (judge verdict: Accepted)

//C - たくさんの数式
//Hiia
#include <stdio.h>
#include <stdlib.h>//rannsuu, zettaichi(absolute)
#include <time.h>
//srand((unsigned int) time(0));
#include <math.h>//koudona keisann
#include <string.h>//hairetsu
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
typedef long long ll;
//#define N 998244353
#define pi 3.1415926535897932384626433832

int gcd(int m,int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
int lcm(int m,int n){if((0==m)||(0==n))return 0;return(m*n/gcd(m, n));}
int fact(int x){if(x==1) return(x);else return(x*fact(x-1));}//Kaijou x!
int ruijou(int base,int exponent){int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

long long int llgcd(long long int m,long long int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
long long int lllcm(long long int m,long long int n){if((0==m)||(0==n))return 0;return(m*n/llgcd(m, n));}
long long int llfact(long long int x){if(x==1) return(x);else return(x*llfact(x-1));}//Kaijou
int llruijou(long long int base,long long int exponent){long long int ii,answer=1;ii=exponent;while(ii!=0){answer=answer*base;--ii;}return answer;}//ruijou base^exponent


long long int llketasuu(long long int x){
  ll jjj=0;
  while(x!=0){
    x/=10;
    jjj++;
  }
  return(jjj);
}

int main()
{
  ll s,ans=0;
  scanf("%lld", &s);
  ll keta=llketasuu(s);
  for(ll i=1;i<=keta;i++){
    if(i==1){
      //printf("%lld;%lld,%lld\n", i,s%10,(s%10)*llruijou(2,keta-1));
      ans+=(s%10)*llruijou(2,keta-1);
    }
    else{
      //ll tmp1=((s/llruijou(10,i-1))%10)*llruijou(2,keta-2)*((9*llruijou(5,i-1) - 1)/4);
      //printf("%lld:%lld,%lld\n", i,(s/llruijou(10,i-1))%10,tmp1);
      ans+=((s/llruijou(10,i-1))%10)*llruijou(2,keta-2)*((9*llruijou(5,i-1) - 1)/4);
    }
  }
  //printf("keta:%lld\n",llketasuu(s));
  //printf("keta:%lld\n",keta);
  printf("%lld\n", ans);

  return 0;
}