JavaEasyNumbers10 min

Check for an Armstrong number

Print true if n equals the sum of its digits each raised to the number of digits. Follow the stated ordering and edge-case conventions exactly. Interviewers use this task to probe algorithm selection, invariants, and boundary handling.

Input

One integer n.

Output

Print true if n equals the sum of its digits each raised to the number of digits.

Constraints

0 <= n <= 2147483647.

Example

Input

Output

true

Example 2 input

2147483647

Output

false

This example exercises a further boundary or ordering case.

Public test cases (2)

Input
```
153
```
Expected output
```
true
```
Input
```
123
```
Expected output
```
false
```

Hints (2)

Identify the state needed to avoid recomputing work for each element.
Count the digits and use integer multiplication for each digit power.

Target complexity: time O(d^2) · space O(1)

Your solution

import java.util.*;
import java.math.BigInteger;

public class Main {
    
    static String armstrong_number(Scanner s) {
        while (s.hasNextLine()) s.nextLine();
        // your code
        return "";
    }
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        System.out.println(armstrong_number(s));
    }
}

Standard input

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Reference solution

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All Java programs

Check for an Armstrong number

Input

Output

Constraints

Example

Public test cases (2)

Your solution

Reference solution

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