Alice and Bob play ping-pong with simplified rules.

During the game, the player serving the ball commences a play. The server strikes the ball then the receiver makes a return by hitting the ball back. Thereafter, the server and receiver must alternately make a return until one of them doesn't make a return.

The one who doesn't make a return loses this play. The winner of the play commences the next play. Alice starts the first play.

Alice has x stamina and Bob has y. To hit the ball (while serving or returning) each player spends 1 stamina, so if they don't have any stamina, they can't return the ball (and lose the play) or can't serve the ball (in this case, the other player serves the ball instead). If both players run out of stamina, the game is over.

Sometimes, it's strategically optimal not to return the ball, lose the current play, but save the stamina. On the contrary, when the server commences a play, they have to hit the ball, if they have some stamina left.

Both Alice and Bob play optimally and want to, firstly, maximize their number of wins and, secondly, minimize the number of wins of their opponent.

Calculate the resulting number of Alice's and Bob's wins.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first and only line of each test case contains two integers x and y (1 ≤ x, y ≤ 10^6) — Alice's and Bob's initial stamina.

Output

For each test case, print two integers — the resulting number of Alice's and Bob's wins, if both of them play optimally.

Example

Input


3
1 1
2 1
1 7


Output


0 1
1 1
0 7

Note

In the first test case, Alice serves the ball and spends 1 stamina. Then Bob returns the ball and also spends 1 stamina. Alice can't return the ball since she has no stamina left and loses the play. Both of them ran out of stamina, so the game is over with 0 Alice's wins and 1 Bob's wins.

In the second test case, Alice serves the ball and spends 1 stamina. Bob decides not to return the ball — he loses the play but saves stamina. Alice, as the winner of the last play, serves the ball in the next play and spends 1 more stamina. This time, Bob returns the ball and spends 1 stamina. Alice doesn't have any stamina left, so she can't return the ball and loses the play. Both of them ran out of stamina, so the game is over with 1 Alice's and 1 Bob's win.

In the third test case, Alice serves the ball and spends 1 stamina. Bob returns the ball and spends 1 stamina. Alice ran out of stamina, so she can't return the ball and loses the play. Bob, as a winner, serves the ball in the next 6 plays. Each time Alice can't return the ball and loses each play. The game is over with 0 Alice's and 7 Bob's wins.
import java.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class CF1455C {
    public static void main(String[] args) throws IOException {
        int t = ri();
        while (t --> 0) {
            int a = rni(), b = ni();
            prln(a - 1 + " " + b);
        }
        close();
    }

    static BufferedReader __in = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter __out = new PrintWriter(new OutputStreamWriter(System.out));
    static StringTokenizer input;

    // references
    // IBIG = 1e9 + 7
    // IRAND ~= 3e8
    // IMAX ~= 2e10
    // LMAX ~= 9e18
    // constants
    static final int IBIG = 1000000007;
    static final int IRAND = 327859546;
    static final int IMAX = 2147483647;
    static final int IMIN = -2147483648;
    static final long LMAX = 9223372036854775807L;
    static final long LMIN = -9223372036854775808L;
    // util
    static int minof(int a, int b, int c) {return min(a, min(b, c));}
    static int minof(int... x) {return x.length == 1 ? x[0] : x.length == 2 ? min(x[0], x[1]) : min(x[0], minstarting(1, x));}
    static int minstarting(int offset, int... x) {assert x.length > 2; return offset == x.length - 2 ? min(x[offset], x[offset + 1]) : min(x[offset], minstarting(offset + 1, x));}
    static long minof(long a, long b, long c) {return min(a, min(b, c));}
    static long minof(long... x) {return x.length == 1 ? x[0] : x.length == 2 ? min(x[0], x[1]) : min(x[0], minstarting(1, x));}
    static long minstarting(int offset, long... x) {assert x.length > 2; return offset == x.length - 2 ? min(x[offset], x[offset + 1]) : min(x[offset], minstarting(offset + 1, x));}
    static int maxof(int a, int b, int c) {return max(a, max(b, c));}
    static int maxof(int... x) {return x.length == 1 ? x[0] : x.length == 2 ? max(x[0], x[1]) : max(x[0], maxstarting(1, x));}
    static int maxstarting(int offset, int... x) {assert x.length > 2; return offset == x.length - 2 ? max(x[offset], x[offset + 1]) : max(x[offset], maxstarting(offset + 1, x));}
    static long maxof(long a, long b, long c) {return max(a, max(b, c));}
    static long maxof(long... x) {return x.length == 1 ? x[0] : x.length == 2 ? max(x[0], x[1]) : max(x[0], maxstarting(1, x));}
    static long maxstarting(int offset, long... x) {assert x.length > 2; return offset == x.length - 2 ? max(x[offset], x[offset + 1]) : max(x[offset], maxstarting(offset + 1, x));}
    static int powi(int a, int b) {if(a == 0) return 0; int ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static long powl(long a, int b) {if(a == 0) return 0; long ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
    static int floori(double d) {return (int)d;}
    static int ceili(double d) {return (int)ceil(d);}
    static long floorl(double d) {return (long)d;}
    static long ceill(double d) {return (long)ceil(d);}
    // input
    static void r() throws IOException {input = new StringTokenizer(__in.readLine());}
    static int ri() throws IOException {return Integer.parseInt(__in.readLine());}
    static long rl() throws IOException {return Long.parseLong(__in.readLine());}
    static int[] ria(int n) throws IOException {int[] a = new int[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Integer.parseInt(input.nextToken()); return a;}
    static long[] rla(int n) throws IOException {long[] a = new long[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Long.parseLong(input.nextToken()); return a;}
    static char[] rcha() throws IOException {return __in.readLine().toCharArray();}
    static String rline() throws IOException {return __in.readLine();}
    static int rni() throws IOException {input = new StringTokenizer(__in.readLine()); return Integer.parseInt(input.nextToken());}
    static int ni() {return Integer.parseInt(input.nextToken());}
    static long rnl() throws IOException {input = new StringTokenizer(__in.readLine()); return Long.parseLong(input.nextToken());}
    static long nl() {return Long.parseLong(input.nextToken());}
    // output
    static void pr(int i) {__out.print(i);}
    static void prln(int i) {__out.println(i);}
    static void pr(long l) {__out.print(l);}
    static void prln(long l) {__out.println(l);}
    static void pr(double d) {__out.print(d);}
    static void prln(double d) {__out.println(d);}
    static void pr(char c) {__out.print(c);}
    static void prln(char c) {__out.println(c);}
    static void pr(char[] s) {__out.print(new String(s));}
    static void prln(char[] s) {__out.println(new String(s));}
    static void pr(String s) {__out.print(s);}
    static void prln(String s) {__out.println(s);}
    static void pr(Object o) {__out.print(o);}
    static void prln(Object o) {__out.println(o);}
    static void prln() {__out.println();}
    static void pryes() {__out.println("yes");}
    static void pry() {__out.println("Yes");}
    static void prY() {__out.println("YES");}
    static void prno() {__out.println("no");}
    static void prn() {__out.println("No");}
    static void prN() {__out.println("NO");}
    static void pryesno(boolean b) {__out.println(b ? "yes" : "no");};
    static void pryn(boolean b) {__out.println(b ? "Yes" : "No");}
    static void prYN(boolean b) {__out.println(b ? "YES" : "NO");}
    static void prln(int... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
    static void prln(long... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
    static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for(int i = 0; i < n; __out.print(iter.next()), __out.print(' '), ++i); __out.println(iter.next());}
    static void flush() {__out.flush();}
    static void close() {__out.close();}}