// #pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define IOS                      \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
#define pb push_back
#define st first
#define nd second
#define PII pair<ll, ll>
#define D(x, y) cout << x << "=" << y << endl;
#define fcout(x, n) cout << fixed << setprecision(x) << n << endl;
#define rep(i, a, b) for (int i = a; i < b; i++)
#define rep2(i, a, b) for (int i = a; i >= b; i--)
// #define int long long
typedef long long ll;
typedef unsigned long long ull;
const ll inf = 0x3f3f3f3f;
const int N = 2e5 + 10;
const int M = 1e9+7;
ll n,m;
ll a[20],ans[N];
ll cnt;
bool bb=0;
ll mpow(ll x,ll y)
{
    ll s=1;
    rep(i,0,y) s*=x; 
    return s;
}
void as(ll l,ll k)
{
    ll s0=0,x=1;
    while(a[x]==0) x++,s0++;
    if(l-x+1<=k||a[k+x]!=9) //k比位数长 || 正常后k位
    {
        ll t=0;
        if(l-x+1<=k) rep2(i,l,x) t=t*10+a[i];
        else if(a[x+k]!=9) rep2(i,k+x-1,x) t=t*10+a[i];
        cout<<mpow(10,k)-t;
        rep(i,0,s0) cout<<0;
        return;
    }
    ll x9=k+x,s9=0;
    while(a[x9+1]==9&&x9<=l-1) x9++;
    rep2(i,x9,x)
    {
        if(a[i]==9) s9++;
        else break;
    }
    if(a[x9-k+1]!=0) //最前面k位
    {
        ll t=0;
        rep2(i,x9,x9-k+1) t=t*10+a[i];
        cout<<mpow(10,k)-t;
        rep(i,0,x9-k+s0) cout<<0;
        return;
    }
    if(s9>=k) //前导9长度大于k
    {
        cout<<1;
        rep(i,0,x9-k) cout<<0;
        return;
    }
    // 前导9长小于k 分段
    ll xx9=x9-k+1;
    while(a[xx9]==0) xx9++;
    ll t=0;
    ll l1=x9-xx9+1,l2=k-l1;
    rep2(i,x9,xx9) t=t*10+a[i];
    cout<<mpow(10,l1)-t;
    while(a[--xx9]==0) cout<<0;
    bb=1;
    as(xx9,l2);
}
void solve()
{
    cnt=0;bb=0;
    string s;cin>>s;
    ll l=s.length();
    rep(i,1,l+1) a[i]=s[l-i]-48;
    ll k;cin>>k;
    if(k==0) //k=0
    {
        ll x=1;
        while(a[x]==9) x++;
        cout<<1;
        rep(i,0,x-1) cout<<0;
        cout<<endl;
        return;
    }
    as(l,k);
    cout<<endl;
}
signed main()
{
    IOS
        // freopen("../.vscode/io/in.txt","r",stdin),freopen("../.vscode/io/out.txt","w",stdout);
        ll T = 1;
    cin >> T;
    while (T--)
        solve();
}
/*
5
1234999 0
323000 2
123000 6
990099 5
990009 5
*/

/*
11
99301093 1
99301093 2
99301093 3
99301093 4
99301093 5
99301093 6
99301093 7
99301093 8
99301093 9
99301093 10
99301093 18
*/