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.

Examples

Input

125


Output

176


Input

9999999999


Output

12656242944
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws IOException {
		Main mainObj = new Main();
		mainObj.solve();
	}

	public void solve() throws IOException {
		FastScanner fs = new FastScanner();
		String s = fs.next();
		System.out.println(dfs(0,0,0,s));
	}
	
	long dfs(long leftNum, long sum, int depth, String s) {
		int nowCharNum = s.charAt(depth) - '0';
		if(depth == s.length() - 1) {
			return sum + leftNum*10 + nowCharNum;
		}
		// +を置かない
		return dfs(leftNum*10+nowCharNum,sum, depth+1, s) + dfs(0, sum + leftNum*10+nowCharNum, depth+1, s);
		
	}

	

	public class FastScanner {

		BufferedReader reader;
		private StringTokenizer st;

		public FastScanner() {
			st = null;
			reader = new BufferedReader(new InputStreamReader(System.in));
		}

		public String next() throws IOException {
			if (st == null || !st.hasMoreElements()) {
				st = new StringTokenizer(reader.readLine());
			}
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			st = null;
			String readLine = null;
			readLine = reader.readLine();
			return readLine;
		}

		public int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		public int[] nextIntArr(int n) throws NumberFormatException, IOException {
			int[] retArr = new int[n];
			for (int i = 0; i < n; i++) {
				retArr[i] = nextInt();
			}
			return retArr;
		}

		public long[] nextLongArr(int n) throws NumberFormatException, IOException {
			long[] retArr = new long[n];
			for (int i = 0; i < n; i++) {
				retArr[i] = nextLong();
			}
			return retArr;
		}

		public void close() throws IOException {
			reader.close();
		}
	}

}
