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 <algorithm>
#include <cstdio>
#include <vector>
#include <math.h>
#include <iomanip>
#include <bitset>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <utility>
#include <set>
using namespace std;
typedef long long int ll;
typedef long double ld;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const long long INF = 1LL << 60;
typedef pair<ll,ll> pairs;
bool pairCompare(const pair<double,ll>& firstElof, const pair<double,ll>& secondElof){
    return firstElof.first < secondElof.first;
}
#define MAX_N 1000100
bool x[MAX_N];
ll num[MAX_N];
ll fibl[MAX_N]={0};
// 四方向への移動ベクトル
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
 
ll fib(ll a){
    if (fibl[a]!=0)return fibl[a];
    if (a==0){
        return 0;
    }else if (a==1){
        return 1;
    }
    return fibl[a]=fib(a-1)+fib(a-2);
}
 
ll eratosthenes(ll n) {
    int p = 0;
    for (ll i=0; i<=n; ++i) x[i] = true;
    x[0] = x[1] = false;
    for(int i=2; i<=n; ++i) {
        if(x[i]) {
            p++;
            for(int j=2*i; j<=n; j+=i) x[j] = false;
        }
        
        num[i] = p;
    }
    return p;
}
 
ll gcd(ll a,ll b){
    if (a%b==0)return(b);
    else return(gcd(b,a%b));
}
 
ll keta(ll N){
    int tmp{};
    while( N > 0 ){
        tmp += ( N % 10 );
        N /= 10;
    }
    N = tmp;
    return N;
}

int main(){
    string s;
    cin >> s;
    ll ans=0;
    for (int bit = 0; bit < (1<<(s.size()-1));bit++){
        ll count=s[0]-'0';
        for (ll i=1;i<s.size();i++){
            if (bit & (1<<(i-1))){
                count=count*10+(s[i]-'0');
            }else{
                ans+=count;
                count=s[i]-'0';
            }
        }
        ans+=count;
    }
    cout << ans << endl;
    return 0;
}