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: Runtime Error)

    /**   Created by: Marcos Portales
          Codeforces User: marcosportales
          Country: Cuba
          Copyright                    */
#include<bits/stdc++.h>
using namespace std;
/// Pragmas:
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline") //Optimization flags
#pragma GCC option("arch=native","tune=native","no-zero-upper") //Enable AVX
#pragma GCC target("avx2") //Enable AVX
/// Macros:
//#define int long long
#define f first
#define s second
#define db(x) cerr << #x << ": " << (x) << '\n';
#define pb push_back
#define lb lower_bound
#define up upper_bound
#define all(x) x.begin() , x.end()
#define rall(x) x.rbegin() , x.rend()
#define enl '\n'
#define vi vector<int>
#define sz(a) int((a).size())
#define rep(i,x,y) for(int i=x;i<=y;i++)
typedef pair<int,int> ii;
typedef long double ld;
typedef unsigned long long ull;
/// Constraints:
const int maxn = 400010;
const int mod = 1000000007;
const ld eps = 1e-9;
const int inf = ((1ll<<31ll)-1ll);
//const int INF = 2000000000000000000ll;
const ld pi = acos(-1);
/// Prime Numbers:
// 2, 173, 179, 181, 197, 311, 331, 737, 1009, 2011, 2027, 3079, 4001, 10037, 10079, 20011, 20089;
// 100003, 100019, 100043, 200003, 200017, 1000003, 1000033, 1000037, 1000081;
// 2000003, 2000029, 2000039, 1000000007, 1000000021, 2000000099;
/// Functions:
#define lg2(x) 31 - __builtin_clz(x)
#define lgx(x,b) ( log(x) / log(b) )
/// Red-Black Tree Template ---------------------------------
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree < long long ,  null_type ,  less<long long> ,  rb_tree_tag ,  tree_order_statistics_node_update > ordered_set;
/// Quick Pow------------------------------------------------
int qpow(int b,int e){
    if( !e ) return 1;
    if( e & 1 ) return qpow(b,e-1) * b % mod;
    int pwur = qpow(b,e>>1);
    return (pwur * pwur) % mod;
}
int modinv(int x){
    return qpow(x,mod-2);
}
/// My Code -------------------------------------------------

int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cout.setf(ios::fixed); cout.precision(0);
    srand(time(NULL));

    //freopen("a.in","r",stdin);
    //freopen("a.in","w",stdout);

    string s;
    cin>>s;
    int n=s.size();
    --n;
    int ans=0;
    rep(i,0,1<<n){
        int last=n;
        rep(j,0,n){
            if(i&(1<<j)){
                int st=last-j-1;
                int num=stoll(s.substr(st,last-st+1));
                ans+=num;
                last=st-1;
            }
        }
        int num=stoll(s.substr(0,last+1));
        ans+=num;
    }

    cout<<ans;

    return 0;
}