Problem: You are given three integers a, b and c.

Find two positive integers x and y (x > 0, y > 0) such that: 

  * the decimal representation of x without leading zeroes consists of a digits; 
  * the decimal representation of y without leading zeroes consists of b digits; 
  * the decimal representation of gcd(x, y) without leading zeroes consists of c digits. 



gcd(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.

Output x and y. If there are multiple answers, output any of them.

Input

The first line contains a single integer t (1 ≤ t ≤ 285) — the number of testcases.

Each of the next t lines contains three integers a, b and c (1 ≤ a, b ≤ 9, 1 ≤ c ≤ min(a, b)) — the required lengths of the numbers.

It can be shown that the answer exists for all testcases under the given constraints.

Additional constraint on the input: all testcases are different.

Output

For each testcase print two positive integers — x and y (x > 0, y > 0) such that 

  * the decimal representation of x without leading zeroes consists of a digits; 
  * the decimal representation of y without leading zeroes consists of b digits; 
  * the decimal representation of gcd(x, y) without leading zeroes consists of c digits. 

Example

Input


4
2 3 1
2 2 2
6 6 2
1 1 1


Output


11 492
13 26
140133 160776
1 1

Note

In the example: 

  1. gcd(11, 492) = 1 
  2. gcd(13, 26) = 13 
  3. gcd(140133, 160776) = 21 
  4. gcd(1, 1) = 1 

Solution (in JAVA):

Solution:
//package Codeforces.EducationalR107;

import java.io.*;
import java.util.*;

public class ProbB {
    final int mod = (int) 1e9 + 7;
    final double eps = 1e-6;
    final double pi = Math.PI;
    final long inf = Long.MAX_VALUE / 2;


    //------------> Solution starts here!!
    @SuppressWarnings("Main Logic")
    void solve() {
        long primes[] = new long[]{1, 1, 97, 997, 9973, 99991, 999983, 9999991, 99999989, 999999937 };
        for (int tc = ii(); tc > 0; tc--) {
            int a = ii(), b = ii(), c = ii();
            long p = primes[c];
            long x = p, y = p;
            while (String.valueOf(x).length() < a) {
                x = x * 2;
            }
            while (String.valueOf(y).length() < b) {
                y = y * 3;
            }
            out.println(x + " " + y);
        }
    }

    //------------> Solution ends here!!
    InputStream obj;
    PrintWriter out;
    String check = "";

    public static void main(String[] args) throws IOException {
        new Thread(null, new Runnable() {
            public void run() {
                try {
                    new ProbB().main1();
                } catch (IOException e) {
                    e.printStackTrace();
                } catch (StackOverflowError e) {
                    System.out.println("RTE");
                }
            }
        }, "1", 1 << 26).start();
    }

    void main1() throws IOException {
        out = new PrintWriter(System.out);
        obj = check.isEmpty() ? System.in : new ByteArrayInputStream(check.getBytes());
        // obj=check.isEmpty() ? new FileInputStream("/home/ankurverma1994/input.txt") : new ByteArrayInputStream(check.getBytes());
        solve();
        out.flush();
        out.close();
    }

    byte inbuffer[] = new byte[1024];
    int lenbuffer = 0, ptrbuffer = 0;

    int readByte() {
        if (lenbuffer == -1) throw new InputMismatchException();
        if (ptrbuffer >= lenbuffer) {
            ptrbuffer = 0;
            try {
                lenbuffer = obj.read(inbuffer);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }
        if (lenbuffer <= 0) return -1;
        return inbuffer[ptrbuffer++];
    }

    String is() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) // when nextLine, (isSpaceChar(b) && b!=' ')
        {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    int ii() {
        int num = 0, b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    long il() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    boolean isSpaceChar(int c) {
        return (!(c >= 33 && c <= 126));
    }

    int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b)) ;
        return b;
    }

    float nf() {
        return Float.parseFloat(is());
    }

    double id() {
        return Double.parseDouble(is());
    }

    char ic() {
        return (char) skip();
    }

    int[] iia(int n) {
        int a[] = new int[n];
        for (int i = 0; i < n; i++) a[i] = ii();
        return a;
    }

    long[] ila(int n) {
        long a[] = new long[n];
        for (int i = 0; i < n; i++) a[i] = il();
        return a;
    }

    String[] isa(int n) {
        String a[] = new String[n];
        for (int i = 0; i < n; i++) a[i] = is();
        return a;
    }

    double[][] idm(int n, int m) {
        double a[][] = new double[n][m];
        for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) a[i][j] = id();
        return a;
    }

    int[][] iim(int n, int m) {
        int a[][] = new int[n][m];
        for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) a[i][j] = ii();
        return a;
    }
}