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)

#ifdef _DEBUG
#include "MyLib.h"
#else
#define main_C main
#include "bits/stdc++.h"
#include <regex>
#define _USE_MATH_DEFINES
#include <math.h>
#define FOR(i,s,e) for (int i = int(s); i < int(e); ++i)
#define REP(i,e) FOR(i,0,e)
#define INF (INT_MAX/2)
#define EPS (1.0e-8)
#define LINF (LONG_MAX/2)
const int MGN = 8;
const int ARY_SZ_MAX = 10000000;
using namespace std;
using ll = long long; using ull = unsigned long long;
using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;
using vl = vector<ll>; using vvl = vector<vl>;
using vd = vector<double>; using vs = vector<string>;
using pii = pair<int, int>; using pll = pair<ll, ll>;
using psi = pair<string, int>;
// functions
vs split(string str, char sep) {
    vs v; stringstream ss(str); string t;
    while (getline(ss, t, sep)) v.push_back(t);
    return v;
}
#endif
string S;
int N;
ll ans;

void calc(vi& use) {
    string t = {S[0]};
    REP(i,N-1) {
        if (use[i]==1){
            t += "+";
        }
        t += S[i + 1];
    }

    vs terms = split(t, '+');
    for(string tm : terms) {
        ans += stoll(tm);
    }
}

void dfs(int n, vi& use) {
    if (n >= N - 1) {
        calc(use);
        return;
    }

    REP(i,2) { // 1:+, 0:nop
        use[n] = i;
        dfs(n + 1, use);
    }
}

int main_C() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> S;
    N = S.size();

    ans = 0;
    vi use(N - 1, 0);
    dfs(0, use);

    cout << ans << endl;

    return 0;
}