The time on the planet Lapituletti goes the same way it goes on Earth but a day lasts h hours and each hour lasts m minutes. The inhabitants of that planet use digital clocks similar to earth ones. Clocks display time in a format HH:MM (the number of hours in decimal is displayed first, then (after the colon) follows the number of minutes in decimal; the number of minutes and hours is written with leading zeros if needed to form a two-digit number). Hours are numbered from 0 to h-1 and minutes are numbered from 0 to m-1. 

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That's how the digits are displayed on the clock. Please note that digit 1 is placed in the middle of its position. 

A standard mirror is in use on the planet Lapituletti. Inhabitants often look at the reflection of the digital clocks in the mirror and feel happy when what you see on the reflected clocks is a valid time (that means that you see valid digits in the reflection and this time can be seen on the normal clocks at some moment of a day).

The image of the clocks in the mirror is reflected against a vertical axis. 

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The reflection is not a valid time.

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The reflection is a valid time with h=24, m = 60. However, for example, if h=10, m=60, then the reflection is not a valid time. 

An inhabitant of the planet Lapituletti begins to look at a mirrored image of the clocks at some time moment s and wants to know the nearest future time moment (which can possibly happen on the next day), when the reflected clock time is valid.

It can be shown that with any h, m, s such a moment exists. If the reflected time is correct at the moment the inhabitant began to look at the clock, that moment is considered the nearest.

You are asked to solve the problem for several test cases.

Input

The first line contains a single integer T (1 ≤ T ≤ 100) — the number of test cases.

The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.

The first line of a test case contains two integers h, m (1 ≤ h, m ≤ 100).

The second line contains the start time s in the described format HH:MM.

Output

For each test case output in a separate line the nearest moment in format HH:MM when the reflected time is correct.

Example

Input


5
24 60
12:21
24 60
23:59
90 80
52:26
1 100
00:01
10 10
04:04


Output


12:21
00:00
52:28
00:00
00:00

Note

In the second test case it is not hard to show that the reflection of 23:59 is incorrect, while the reflection of the moment 00:00 on the next day is correct. 

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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.PrintStream;
import java.io.PrintWriter;
public class Main {
 
	
	/* 10^(7) = 1s.
	 * ceilVal = (a+b-1) / b */
	
	static final int mod = 1000000007;
	static final long temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	static class Pair implements Comparable<Pair> 
	{
		int first, second;
		public Pair(int aa, int bb)
		{
			first = aa; second = bb;
		}
		public int compareTo(Pair o)
		{
			if(this.second < o.second) return -1;
			if(this.second > o.second) return +1;
			return this.first - o.first;
		}
	}
	
	public static class DSU
	{
		int[] parent;
		int[] rank; //Size of the trees is used as the rank
		public DSU(int n)
		{
			parent = new int[n];
			rank = new int[n];
			Arrays.fill(parent, -1);
			Arrays.fill(rank, 1);
		}
		
		public int find(int i) //finding through path compression
		{
			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
		}
		
		public boolean union(int a, int b) //Union Find by Rank 
		{
			a = find(a);
			b = find(b);
			
			if(a == b) return false; //if they are already connected we exit by returning false.
			
			// if a's parent is less than b's parent
			if(rank[a] < rank[b])
			{
				//then move a under b
				parent[a] = b;
			}
			//else if rank of j's parent is less than i's parent
			else if(rank[a] > rank[b])
			{
				//then move b under a
				parent[b] = a;
			}
			//if both have the same rank.
			else
			{
				//move a under b (it doesnt matter if its the other way around.
				parent[b] = a;
				rank[a] = 1 + rank[a];
			}
			return true;
		}	
	}

	static class Reader {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
	public static boolean isPrime(long n) 
	{
		if(n == 1)
		{
			return false;
		}
		for(int i = 2;i*i<=n;i++)
		{
			if(n%i == 0)
			{
				return false;
			}
		}
		return true;
}
	
	public static ArrayList<Integer> SieveList(int n)
	{
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime, true);
		
		ArrayList<Integer> l = new ArrayList<>();
		for (int p=2; p*p<=n; p++) 
		{ 
			if (prime[p] == true) 
			{ 	
			    for(int i=p*p; i<=n; i += p) 
				{
				    prime[i] = false; 
				}	
			} 
		} 
 
		for (int p=2; p<=n; p++) 
		{
		    if (prime[p] == true)
		    {
		       l.add(p); 
		    }
		}
		
		return l;
	}
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
		public static int lcm(int a, int b)
	    {
	        return (a / gcd(a, b)) * b;
	    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
		public static long LongLCM(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
		public static long phi(long n)  //euler totient function
	    { 
	       long ans = n;
//	       for(long i = 2;i*i<=n;i++)
//	       {
//	    	   if(n%i == 0)
//	    	   {
//	    		   while(n%i == 0) n/=i;
//	    		   ans -= (ans/i);
//	    	   }
//	       }
//	       
//	       if(n > 1)
//	       {
//	    	   ans -= (ans/n);
//	       }
	       
	       for(long i = 2;i<=n;i++)
	       {
	    	   if(isPrime(i))
	    	   {
	    		   ans -= (ans/i);
	    	   }
	       }
	       
	       return ans;
	       
	    }
	
	public static int fastPow(int x, int n) 
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	   public static long modPow(long x, long y, long p)
	    {
	        long res = 1;
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCrModP(long n, long r,long p)
	    {
	    	if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long fact(long n) {
	    	  long[] fac = new long[(int)(n) + 1];
		        fac[0] = 1;
		 
		        for (long i = 1; i <= n; i++)
		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
		        
		        return fac[(int)(n)];
		}
	    
	    public static long nCr(int n, int k) 
	    {
			long ans = 1;
			for(int i = 0;i<k;i++)
			{
				ans *= (n-i);
				ans /= (i+1);
			}
			return ans;
		}
 
		public static void Sort(int[] a) {
			List<Integer> l = new ArrayList<>();
			for (int i : a) l.add(i);
			Collections.sort(l);
//			Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		public static void ssort(char[] a)
		{
			List<Character> l = new ArrayList<>();
			for (char i : a) l.add(i);
			Collections.sort(l);
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		//Modular Operations for Addition and Multiplication. 
		   public static long perfomMod(long x)
		   {
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		  
		   public static long subMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)-perfomMod(b));
		   }
		   
		   public static long mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		   
		   /*
		    * 
		    * >=         <=
		   0 1 2 3 4 5 6 7  
		   5 5 5 6 6 6 7 7
		   lower_bound for 6 at index 3 (>=)
		   upper_bound for 6 at index 6(To get six reduce by one) (<=)
		   */
		
		   public static int LowerBound(int a[], int x) 
		   {
				  int l=-1,r=a.length;
				  while(l+1<r) {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
			}
		
		   
			public static int UpperBound(int a[], int x) 
			{
			    int l=-1, r=a.length;
			    while(l+1<r) {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
			}

		
	public static void main(String[] args) throws Exception 
	{
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);
		
		int[] a = {0, 1, 5, -1, -1, 2, -1, -1, 8, -1};
		int t = sc.nextInt();
	f:	while(t-- > 0)
		{
			int inpH  = sc.nextInt(), inpM = sc.nextInt();
			String str = sc.next();
			int x = Integer.parseInt(str.substring(0,2));
			int y = Integer.parseInt(str.substring(3,5));
			
			int s = x * inpM + y;
			for(int i = s;;i++)
			{
				if(i >= inpH*inpM) i %= inpH*inpM;
				
				int h = i / inpM, m = i % inpM;
				if(a[h%10] == -1)continue;
				if(a[h/10] == -1)continue;
				if(a[m%10] == -1)continue;
				if(a[m/10] == -1)continue;
				
				int mm = a[h%10] * 10 + a[h/10];
				int hh = a[m%10] * 10 + a[m/10];
				
				if(hh < inpH && mm < inpM)
				{
					out.println(String.format("%02d:%02d", h, m));
					continue f;
				}
				
			}
			
		}
		
		fout.close();
		
	}
}
