/**
 *  - Prakhar Mishra
 *  - The force is strong with this code.
**/

#include <bits/stdc++.h>
using namespace std;
using ll=long long;

void solve(int tt)
{
    // taking inputs

    string s;
    cin>>s;

    ll k;
    cin>>k;

    // if k==0, no operations, directly output
    
    if(k==0)
    {
        cout<<s<<endl;
        return;
    }

    // find the minimum digit from the first (k+1) digits of x excluding '0' as leading zero not allowed.
    
    ll mn=10;
    ll ind=-1;
    for(ll i=0;i<=k;i++)
    {
        if(s[i]!='0' && s[i]-'0'<mn)
        {
            mn=s[i]-'0';
            ind=i;
        }
    }

    // remove everything before the smallest digit in x and reduce k accordingly.
    // add to answer the first digit.

    string ans="";
    ans+=s[ind];

    k=k-ind;

    // now we repeat this process again till k>0, but we will include '0' also in calculating minimum this time
    
    // range to choose next digit from =[ind+1,ind+1+k]
    ll p1=ind+1;
    ll p2=p1+k;

    while(k>0)
    {
        // finding minimum digit

        ll mn=10;
        ll ind=-1;
        for(ll i=p1;i<=p2;i++)
        {
            if(s[i]-'0'<mn)
            {
                mn=s[i]-'0';
                ind=i;
            }     
        }

        // remove everything before the smallest digit in range and reduce k accordingly.
        // and add to answer the minimum digit. 
        
        for(ll i=p1;i<ind;i++)
        {
            k--;
        }
        ans+=s[ind];

        // set the range to choose from for next digit
        p1=ind+1;
        p2=p1+k;
    }  

    // lastly include everything left after operations are done
    for(ll i=p1;i<s.size();i++)
        ans+=s[i];

    // output
    cout<<ans<<endl;
}

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    int T=1;
    cin >> T;

    for(int tt=1; tt<=T; tt++)
    {
        solve(tt);
    }
    return 0;
}