#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <cstring>
using namespace std;

const int MAXN = 3e6 + 5, N = 3e6 + 5;
bool _stmer;

struct bign
{  
    int len, s[MAXN];  
    bign ()  //初始化
    {  
        memset(s, 0, sizeof(s));  
        len = 1;  
    }  
    bign (int num) { *this = num; }  
    bign (const char *num) { *this = num; }  //让this指针指向当前字符串
    bign operator = (const int num)  
    {  
        char s[MAXN];  
        sprintf(s, "%d", num);  //sprintf函数将整型映到字符串中
        *this = s;  
        return *this;  //再将字符串转到下面字符串转化的函数中
    }  
    bign operator = (const char *num)  
    {  
        for(int i = 0; num[i] == '0'; num++) ;  //去前导0  
        len = strlen(num);  
        for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0'; //反着存
        return *this;
    }  
    bign operator + (const bign &b) const //对应位相加，最为简单
    {  
        bign c;  
        c.len = 0;  
        for(int i = 0, g = 0; g || i < max(len, b.len); i++)  
        {  
            int x = g;  
            if(i < len) x += s[i];  
            if(i < b.len) x += b.s[i];  
            c.s[c.len++] = x % 10;  //关于加法进位
            g = x / 10;  
        }  
        return c;  
    }  
    bign operator += (const bign &b)  //如上文所说，此类运算符皆如此重载
    {  
        *this = *this + b;  
        return *this;  
    }  
    void clean()  //由于接下来的运算不能确定结果的长度，先大而估之然后再查
    {  
        while(len > 1 && !s[len-1]) len--;  //首位部分‘0’故删除该部分长度
    }   
    bign operator * (const bign &b) //乘法重载在于列竖式，再将竖式中的数转为抽象，即可看出运算法则。
    {  
        bign c;  
        c.len = len + b.len;  
        for(int i = 0; i < len; i++)  
        {  
            for(int j = 0; j < b.len; j++)  
            {  
                c.s[i+j] += s[i] * b.s[j];//不妨列个竖式看一看
            }  
        }  
        for(int i = 0; i < c.len; i++) //关于进位，与加法意同 
        {  
            c.s[i+1] += c.s[i]/10;
            c.s[i] %= 10;
        }  
        c.clean();  //我们估的位数是a+b的长度和，但可能比它小（1*1 = 1）
        return c;  
    }  
    bign operator *= (const bign &b)  
    {  
        *this = *this * b;  
        return *this;  
    }  
    bign operator - (const bign &b)  //对应位相减，加法的进位改为借1
    {  //不考虑负数
        bign c;  
        c.len = 0;  
        for(int i = 0, g = 0; i < len; i++)  
        {  
            int x = s[i] - g;  
            if(i < b.len) x -= b.s[i];  //可能长度不等
            if(x >= 0) g = 0;  //是否向上移位借1
            else  
            {  
                g = 1;  
                x += 10;  
            }  
            c.s[c.len++] = x;  
        }  
        c.clean();  
        return c;   
    }  
    bign operator -= (const bign &b)  
    {  
        *this = *this - b;  
        return *this;  
    }  
    bign operator / (const bign &b)  //运用除是减的本质，不停地减，直到小于被减数
    {  
        bign c, f = 0; //可能会在使用减法时出现高精度运算 
        for(int i = len-1; i >= 0; i--)  //正常顺序，从最高位开始
        {  
            f = f*10;  //上面位的剩余到下一位*10
            f.s[0] = s[i];  //加上当前位
            while(f >= b)  
            {  
                f -= b;  
                c.s[i]++;  
            }  
        }  
        c.len = len;  //估最长位
        c.clean();  
        return c;  
    }  
    bign operator /= (const bign &b)  
    {  
        *this  = *this / b;  
        return *this;  
    }  
    bign operator % (const bign &b)  //取模就是除完剩下的
    {  
        bign r = *this / b;  
        r = *this - r*b;  
        r.clean();
        return r;  
    }  
    bign operator %= (const bign &b)  
    {  
        *this = *this % b;  
        return *this;  
    }  
    bool operator < (const bign &b) //字符串比较原理 
    {  
        if(len != b.len) return len < b.len;  
        for(int i = len-1; i != -1; i--)  
        {  
            if(s[i] != b.s[i]) return s[i] < b.s[i];  
        }  
        return false;  
    }  
    bool operator > (const bign &b)  //同理
    {  
        if(len != b.len) return len > b.len;  
        for(int i = len-1; i != -1; i--)  
        {  
            if(s[i] != b.s[i]) return s[i] > b.s[i];  
        }  
        return false;  
    }  
    bool operator == (const bign &b)  
    {  
        return !(*this > b) && !(*this < b);  
    }  
    bool operator != (const bign &b)  
    {  
        return !(*this == b);  
    }  
    bool operator <= (const bign &b)  
    {  
        return *this < b || *this == b;  
    }  
    bool operator >= (const bign &b)  
    {  
        return *this > b || *this == b;  
    }  
    string str() const  //将结果转化为字符串（用于输出）
    {  
        string res = "";  
        for(int i = 0; i < len; i++) res = char(s[i]+'0')+res;  
        return res;  
    }  
};  

istream& operator >> (istream &in, bign &x) //重载输入流 
{  
    string s;  
    in >> s;  
    x = s.c_str();  //string转化为char[]
    return in;  
}  

ostream& operator << (ostream &out, const bign &x)  //重载输出流
{  
    out << x.str();  
    return out;  
}

int strbuf[N];

void solve() {
    bign n, m;
    cin >> n >> m;
    bign ans = n + m - (n % m);
    /* cout << ans << endl; */
    bign res1 = n + 1;
    bool flg1 = 0;
    for (int i = 0; i + m.len < res1.len; i++) {
        bool flg = 0;
        for (int j = m.len - 1; ~j; j--)
            if (res1.s[i + j] != m.s[j]) { flg = 1; break; }
        if (flg) continue;
        flg1 = 1; break;
    }
    if (flg1) { return (void)(cout << res1 << endl); }

    bign res2 = 0; int pos2 = -1;
    for (int i = 0; i + m.len < n.len; i++) {
        bool flg = 0;
        for (int j = m.len - 1; ~j; j--) {
            if (res1.s[i + j] == m.s[j]) continue;
            if (res1.s[i + j] > m.s[j]) { flg = 1; break; }
            if (res1.s[i + j] < m.s[j]) { flg = 0; break; }
        }
        if (flg) continue;
        pos2 = i;
    }
    if (~pos2) {
        bign tp = 0;
        for (int i = pos2 + m.len; i < n.len; i++)
            tp.s[tp.len++] = n.s[i];
        tp = tp;
        for (int i = 0; i < pos2; i++)
            res2.s[res2.len++] = 0;
        int pos3 = pos2 + m.len;
        for (int i = pos2; i < pos3; i++)
            res2.s[res2.len++] = m.s[i - pos2];
        for (int i = pos3; i < pos3 + tp.len; i++)
            res2.s[res2.len++] = tp.s[i - pos3];
    }

    bign res3 = 0;
    for (int i = 0; i < m.len; i++)
        res3.s[res3.len++] = m.s[i];
    bign tp = 0;
    for (int i = m.len; i < n.len; i++)
        tp.s[tp.len++] = n.s[i];
    tp = tp + 1;
    for (int i = m.len; i < m.len + tp.len; i++)
        res3.s[res3.len++] = tp.s[i - m.len];
    if (res2.len && res2 < res3)
        res3 = res2;
    if (res3 < ans) cout << res3 << endl;
    else cout << ans << endl;
}

bool _edmer;
int main() {
    cerr << (&_stmer - &_edmer) / 1024.0 / 1024.0 << "MB\n";
    int T; cin >> T;
    while (T--) solve();
    return 0;
}