import sys

# sys.setrecursionlimit(200005)
# sys.set_int_max_str_digits(200005)
int1 = lambda x: int(x)-1
pDB = lambda *x: print(*x, end="\n", file=sys.stderr)
p2D = lambda x: print(*x, sep="\n", end="\n\n", file=sys.stderr)
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()

dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
# inf = -1-(-1 << 31)
inf = -1-(-1 << 62)

# md = 10**9+7
md = 998244353

from math import gcd

# 拡張ユークリッドの互除法(非再帰)
# ax+by=gcd(a,b)となるx,yとgcd(a,b)が返る
# extgcd(a,mod)でg=1のときx=(aの逆元)
# ax≡b (mod m)を解くためにaの逆元を求めるときは
# gcd(a,b,m)でa,b,mを割ってからやること
# そして、a･a^-1==1か、確認すること

def extgcd(a, b):
    qq = []
    while b: a, b, _ = b, a%b, qq.append(a//b)
    x = y = 1
    for q in qq[::-1]: x, y = y, x-q*y
    if a < 0: return -x, -y, -a
    return x, y, a

def eq(a,b,m):
    g=gcd(gcd(a,b),m)
    a,b,m=a//g,b//g,m//g
    inv,_,_=extgcd(a,m)
    if a*inv%m==1:return b*inv%m
    return -1

def solve():
    s,m=SI().split()
    m=int(m)
    l=len(s)
    x=int(s)
    a=pow(10,l,m)
    for k in range(l,19):
        d=k-2*l
        if d<0:
            d=-d
            if s[-d:]==s[:d] and int(s+s[d:])%m==0:
                print(s+s[d:])
                return
        else:
            xx=(10**(d+l)+1)*x
            b=-xx%m
            z=eq(a,b,m)
            # print(k,a,xx,z)
            if z!=-1 and z<10**d:
                print(xx+z*10**l)
                return
    print(-1)

for _ in range(II()):solve()