Your friend Jeff Zebos has been trying to run his new online company, but it's not going very well. He's not getting a lot of sales on his website which he decided to call Azamon. His big problem, you think, is that he's not ranking high enough on the search engines. If only he could rename his products to have better names than his competitors, then he'll be at the top of the search results and will be a millionaire.

After doing some research, you find out that search engines only sort their results lexicographically. If your friend could rename his products to lexicographically smaller strings than his competitor's, then he'll be at the top of the rankings!

To make your strategy less obvious to his competitors, you decide to swap no more than two letters of the product names.

Please help Jeff to find improved names for his products that are lexicographically smaller than his competitor's!

Given the string s representing Jeff's product name and the string c representing his competitor's product name, find a way to swap at most one pair of characters in s (that is, find two distinct indices i and j and swap s_i and s_j) such that the resulting new name becomes strictly lexicographically smaller than c, or determine that it is impossible.

Note: String a is strictly lexicographically smaller than string b if and only if one of the following holds:

  * a is a proper prefix of b, that is, a is a prefix of b such that a ≠ b; 
  * There exists an integer 1 ≤ i ≤ min{(|a|, |b|)} such that a_i < b_i and a_j = b_j for 1 ≤ j < i. 

Input

The first line of input contains a single integer t (1 ≤ t ≤ 1500) denoting the number of test cases. The next lines contain descriptions of the test cases.

Each test case consists of a single line containing two space-separated strings s and c (2 ≤ |s| ≤ 5000, 1 ≤ |c| ≤ 5000). The strings s and c consists of uppercase English letters.

It is guaranteed that the sum of |s| in the input is at most 5000 and the sum of the |c| in the input is at most 5000.

Output

For each test case, output a single line containing a single string, which is either

  * the new name which is obtained after swapping no more than one pair of characters that is strictly lexicographically smaller than c. In case there are many possible such strings, you can output any of them; 
  * three dashes (the string "---" without quotes) if it is impossible. 

Example

Input


3
AZAMON APPLE
AZAMON AAAAAAAAAAALIBABA
APPLE BANANA


Output


AMAZON
---
APPLE

Note

In the first test case, it is possible to swap the second and the fourth letters of the string and the resulting string "AMAZON" is lexicographically smaller than "APPLE".

It is impossible to improve the product's name in the second test case and satisfy all conditions.

In the third test case, it is possible not to swap a pair of characters. The name "APPLE" is lexicographically smaller than "BANANA". Note that there are other valid answers, e.g., "APPEL". 
import java.io.*;
import java.util.*;
import java.lang.*;
import java.math.*;
import java.text.DecimalFormat;
import java.lang.reflect.Array;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
public class B{
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	//static long MOD = (long)(1e9+7);
	static long MOD = 998244353;
	static long MOD2 = MOD*MOD;
	static FastReader sc = new FastReader();
	static int pInf = Integer.MAX_VALUE;
	static int nInf = Integer.MIN_VALUE;
	public static void main(String[] args){
		int test = 1;
		test = sc.nextInt();
		while(test-->0)solve();
		out.flush();
		out.close();
	}
	public static void solve() {
		String s = sc.next();
		char[] t = sc.next().toCharArray();
		char[] c = s.toCharArray();
		char[] p = s.toCharArray();
		Arrays.sort(p);
		for(int i = 0; i < s.length(); i++) {
			if(c[i]!=p[i]) {
				for(int j = c.length-1; j>i; j--) {
					if(s.charAt(j)==p[i]) {
						char temp = c[i];
						c[i] = c[j];
						c[j] = temp;
						break;
					}
				}
				break;
			}
		}
		boolean ans = false;
		int count = 0;
		//out.println(new String(c));
		for(int i = 0; i < Math.min(t.length, c.length); i++) {
			if(c[i]!=t[i]) {
				if(c[i]>t[i]) {
					ans = true;
				}
				break;
			}
			else {
				count++;
			}
		}
		if(ans) {
			out.println("---");
			return;
		}
		if(t.length==count) {
			out.println("---");
			return;
		}
		else {
			out.println(new String(c));
		}
	}
	public static int lowerBound(int key, int[] a) {
		int s = 0;
		int e = a.length-1;
		if(e==-1) {
			return 0;
		}
		int ans = -1;
		while(s<=e) {
			int m = s+(e-s)/2;
			if(a[m]>=key) {
				ans = m;
				e = m-1;
			}
			else {
				s = m+1;
			}
		}
		return ans==-1?s:ans;
	}
	public static int upperBound(int key, int[] a) {
		int s = 0;
		int e = a.length-1;
		if(e==-1) {
			return 0;
		}
		int ans = -1;
		while(s<=e) {
			int m = s+(e-s)/2;
			if(a[m]>key) {
				ans = m;
				e = m-1;
			}
			else {
				s = m+1;
			}
		}
		return ans==-1?s:ans;
	}
	public static long c2(long n) {
		if((n&1)==0) {
			return mul(n/2, n-1);
		}
		else {
			return mul(n, (n-1)/2);
		}
	}
	public static long mul(long a, long b){
	    return ((a%MOD)*(b%MOD))%MOD;
	}
	public static long add(long a, long b){
	    return ((a%MOD)+(b%MOD))%MOD;
	}
	public static long sub(long a, long b){
	    return ((a%MOD)-(b%MOD))%MOD;
	}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//Integer.lowestOneBit(i)   Equals k where k is the position of the first one in the binary
//Integer.highestOneBit(i)  Equals k where k is the position of the last one in the binary
//Integer.bitCount(i)       returns the number of one-bits
//Collections.sort(A,(p1,p2)->(int)(p2.x-p1.x)) To sort ArrayList in descending order wrt values of x.
//            Arrays.parallelSort(a,new Comparator<TPair>() {
//      		public int compare(TPair a,TPair b) {
//        			if(a.y==b.y) return a.x-b.x;
//        			return b.y-a.y;
//        		}
//        	});
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//PrimeFactors    
    public static ArrayList<Long> primeFactors(long n) {
        ArrayList<Long> arr = new ArrayList<>();
        if (n % 2 == 0)
            arr.add((long) 2);
        while (n % 2 == 0)
            n /= 2;
        for (long i = 3; i <= Math.sqrt(n); i += 2) {
            int flag = 0;
            while (n % i == 0) {
                n /= i;
                flag = 1;
            }
            if (flag == 1)
                arr.add(i);
        }
        if (n > 2)
            arr.add(n);
        return arr;
    }
//Pair Class
	static class Pair implements Comparable<Pair>{
		int x;
		int y;
		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
		@Override
		public int compareTo(Pair o) {
			if(this.x==o.x){
			    return -1*(this.y-o.y);
			}
			return (this.x-o.x);
		}
	}
	static class TPair implements Comparable<TPair>{
		int l;
		int r;
		int index;
		public TPair(int l, int r, int index) {
			this.l = l;
			this.r = r;
			this.index = index;
		}
		@Override
		public int compareTo(TPair o) {
			if(this.l==o.l){
			    return -1*(this.r-o.r);
			}
			return -1*(this.l-o.l);
		}
	}
//nCr
	static long ncr(long n, long k) {
        long ret = 1;
        for (long x = n; x > n - k; x--) {
            ret *= x;
            ret /= (n - x + 1);
        }
 
        return ret;
    }
	static long finextDoubleMMI_fermat(long n,int M)
    {
        return fastExpo(n,M-2);
    }
 
    static long nCrModPFermat(int n, int r, int p) 
    { 
        if (r == 0) 
            return 1; 
        long[] fac = new long[n+1]; 
        fac[0] = 1;           
        for (int i = 1 ;i <= n; i++) 
            fac[i] = fac[i-1] * i % p;       
        return (fac[n]* finextDoubleMMI_fermat(fac[r], p)% p * finextDoubleMMI_fermat(fac[n-r], p) % p) % p; 
    }
//Kadane's Algorithm    
    static long maxSubArraySum(ArrayList<Long> a) { 
    	if(a.size()==0) {
    		return 0;
    	}
    	long max_so_far = a.get(0); 
    	long curr_max = a.get(0); 
    	for (int i = 1; i < a.size(); i++) { 
           curr_max = Math.max(a.get(i), curr_max+a.get(i)); 
           max_so_far = Math.max(max_so_far, curr_max); 
    	} 
    	return max_so_far; 
    } 
//Shuffle Sort
    static final Random random=new Random();
	static void ruffleSort(int[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n), temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
//Merge Sort
	static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        long L[] = new long [n1];
        long R[] = new long [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
//Brian Kernighans Algorithm
    static long countSetBits(long n){
        if(n==0) return 0;
        return 1+countSetBits(n&(n-1));
    }
//Factorial
    static long factorial(long n){
        if(n==1) return 1;
        if(n==2) return 2;
        if(n==3) return 6;
        return n*factorial(n-1);
    }
//Euclidean Algorithm
    static long gcd(long A,long B){
        if(B==0) return A;
        return gcd(B,A%B);
    }
//Modular Exponentiation
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
    }
//AKS Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i<=Math.sqrt(n);i+=6)
            if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
//Reverse an array
    static <T> void reverse(T arr[],int l,int r){
    	Collections.reverse(Arrays.asList(arr).subList(l, r));
    }
  //Print array
    static void print1d(int arr[]) {
    	out.println(Arrays.toString(arr));
    }
    static void print2d(int arr[][]) {
    	for(int a[]: arr) out.println(Arrays.toString(a));
    }
//Sieve of eratosthenes
    static int[] findPrimes(int n){
        boolean isPrime[]=new boolean[n+1];
        ArrayList<Integer> a=new ArrayList<>();
        int result[];
        Arrays.fill(isPrime,true);
        isPrime[0]=false;
        isPrime[1]=false;
        for(int i=2;i*i<=n;++i){
            if(isPrime[i]==true){
                for(int j=i*i;j<=n;j+=i) isPrime[j]=false;
            }
        }
        for(int i=0;i<=n;i++) if(isPrime[i]==true) a.add(i);
        result=new int[a.size()];
        for(int i=0;i<a.size();i++) result[i]=a.get(i);
        return result;
        
    }
//Indivisual factors of all nos till n
    static ArrayList<Integer>[] indiFactors(int n){
    	ArrayList<Integer>[] A = new ArrayList[n+1];
    	for(int i = 0; i <= n; i++) {
    		A[i] = new ArrayList<>();
    	}
    	int[] sieve = new int[n+1];
    	for(int i=2;i<=n;i++) {
       		if(sieve[i]==0) {
       			for(int j=i;j<=n;j+=i) if(sieve[j]==0) {
       				//sieve[j]=i;
       				A[j].add(i);
       			}
       		}
       	}
    	return A;
    }
//Segmented Sieve
    static boolean[] segmentedSieve(long l, long r){
    	boolean[] segSieve = new boolean[(int)(r-l+1)];
    	Arrays.fill(segSieve, true);
    	int[] prePrimes = findPrimes((int)Math.sqrt(r));
    	for(int p:prePrimes) {
    		long low = (l/p)*p;
    		if(low < l) {
    			low += p;
    		}
    		if(low == p) {
    			low += p;
    		}
    		for(long j = low; j<= r; j += p) {
    			segSieve[(int) (j-l)] = false;
    		}
    	}
    	if(l==1) {
    		segSieve[0] = false;
    	}
    	return segSieve;
    }
//Euler Totent function
    static long countCoprimes(long n){
        ArrayList<Long> prime_factors=new ArrayList<>();
        long x=n,flag=0;
        while(x%2==0){
            if(flag==0) prime_factors.add(2L);
            flag=1;
            x/=2;
        }
        for(long i=3;i*i<=x;i+=2){
            flag=0;
            while(x%i==0){
                if(flag==0) prime_factors.add(i);
                flag=1;
                x/=i;
            }
        }
        if(x>2) prime_factors.add(x);
        double ans=(double)n;
        for(Long p:prime_factors){
            ans*=(1.0-(Double)1.0/p);
        }
        return (long)ans;
    }
	static long modulo = (long)1e9+7;
	public static long modinv(long x){
	    return modpow(x, modulo-2);
	}
	public static long modpow(long a, long b){
	    if(b==0){
	        return 1;
	    }
	    long x = modpow(a, b/2);
	    x = (x*x)%modulo;
	    if(b%2==1){
	        return (x*a)%modulo;
	    }
	    return x;
	}
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception r) {
					r.printStackTrace();
				}
			}
			return st.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception r) {
				r.printStackTrace();
			}
			return str;
		}
	}
}