You are given a string s consisting of lowercase English letters and a number k. Let's call a string consisting of lowercase English letters beautiful if the number of occurrences of each letter in that string is divisible by k. You are asked to find the lexicographically smallest beautiful string of length n, which is lexicographically greater or equal to string s. If such a string does not exist, output -1.

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

  * a is a prefix of b, but a ≠ b; 
  * in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b. 

Input

The first line contains a single integer T (1 ≤ T ≤ 10 000) — the number of test cases.

The next 2 ⋅ T lines contain the description of test cases. The description of each test case consists of two lines.

The first line of the description contains two integers n and k (1 ≤ k ≤ n ≤ 10^5) — the length of string s and number k respectively.

The second line contains string s consisting of lowercase English letters.

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case output in a separate line lexicographically smallest beautiful string of length n, which is greater or equal to string s, or -1 if such a string does not exist.

Example

Input


4
4 2
abcd
3 1
abc
4 3
aaaa
9 3
abaabaaaa


Output


acac
abc
-1
abaabaaab

Note

In the first test case "acac" is greater than or equal to s, and each letter appears 2 or 0 times in it, so it is beautiful.

In the second test case each letter appears 0 or 1 times in s, so s itself is the answer.

We can show that there is no suitable string in the third test case.

In the fourth test case each letter appears 0, 3, or 6 times in "abaabaaab". All these integers are divisible by 3.
# DEFINING SOME GOOD STUFF
from math import *
import threading
import sys
from collections import defaultdict
from pprint import pprint
sys.setrecursionlimit(300000)
# threading.stack_size(10**8)
'''
-> if you are increasing recursionlimit then remember submitting using python3 rather pypy3
-> sometimes increasing stack size don't work locally but it will work on CF
'''

mod = 10 ** 9
inf = 10 ** 15
yes = 'YES'
no = 'NO'

# ------------------------------FASTIO----------------------------
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")



# _______________________________________________________________#

def npr(n, r):
    return factorial(n) // factorial(n - r) if n >= r else 0


def ncr(n, r):
    return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0


def lower_bound(li, num):
    answer = -1
    start = 0
    end = len(li) - 1

    while (start <= end):
        middle = (end + start) // 2
        if li[middle] >= num:
            answer = middle
            end = middle - 1
        else:
            start = middle + 1
    return answer  # min index where x is not less than num


def upper_bound(li, num):
    answer = -1
    start = 0
    end = len(li) - 1

    while (start <= end):
        middle = (end + start) // 2

        if li[middle] <= num:
            answer = middle
            start = middle + 1

        else:
            end = middle - 1
    return answer  # max index where x is not greater than num


def abs(x):
    return x if x >= 0 else -x


def binary_search(li, val):
    # print(lb, ub, li)
    ans = -1
    lb = 0
    ub = len(li) - 1
    while (lb <= ub):
        mid = (lb + ub) // 2
        # print('mid is',mid, li[mid])
        if li[mid] > val:
            ub = mid - 1
        elif val > li[mid]:
            lb = mid + 1
        else:
            ans = mid  # return index
            break
    return ans


def kadane(x):  # maximum sum contiguous subarray
    sum_so_far = 0
    current_sum = 0
    for i in x:
        current_sum += i
        if current_sum < 0:
            current_sum = 0
        else:
            sum_so_far = max(sum_so_far, current_sum)
    return sum_so_far


def pref(li):
    pref_sum = [0]
    for i in li:
        pref_sum.append(pref_sum[-1] + i)
    return pref_sum


def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    li = []
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1

    for p in range(2, len(prime)):
        if prime[p]:
            li.append(p)
    return li


def primefactors(n):
    factors = []
    while (n % 2 == 0):
        factors.append(2)
        n //= 2
    for i in range(3, int(sqrt(n)) + 1, 2):  # only odd factors left
        while n % i == 0:
            factors.append(i)
            n //= i
    if n > 2:  # incase of prime
        factors.append(n)
    return factors


def prod(li):
    ans = 1
    for i in li:
        ans *= i
    return ans


def dist(a, b):
    d = abs(a[1] - b[1]) + abs(a[2] - b[2])
    return d


def power_of_n(x, n):
    cnt = 0
    while (x % n == 0):
        cnt += 1
        x //= n
    return cnt

def get(x, k):
    return (k - x%k)%k


# _______________________________________________________________#


# def main():
# for _ in range(1):
for _ in range(int(input()) if True else 1):
    # n = int(input())
    n, k = map(int, input().split())
    # a = list(map(int, input().split(':')))
    # b = list(map(int, input().split()))
    # c = list(map(int, input().split()))
    # s = list(input())
    s = input()
    cnt = [0]*26
    for i in s:
        cnt[ord(i)-97] += 1
    to_change = 0
    for i in range(26):
        to_change += get(cnt[i], k)  # find the number of more occurances you need to make it k divisible

    if to_change == 0:
        print(s)
        continue
    if n%k != 0:
        print(-1)
        continue

    flag = 1
    for i in range(n-1, -1, -1):
        if flag == 0: break

        to_change -= get(cnt[ord(s[i]) - 97], k)
        cnt[ord(s[i]) - 97] -= 1
        to_change += get(cnt[ord(s[i]) - 97], k)  # taking out last element one by one to find the part that has to be kept as it is

        for j in range(ord(s[i])-96, 26):
            last_change = to_change
            to_change -= get(cnt[j], k)
            cnt[j] += 1
            to_change += get(cnt[j], k)  # trying every element just after the window of elements that you kept the same

            if i + to_change + 1 <= n:
                print(s[:i]+chr(j+97), end='')

                to_add = []
                for m in range(26):
                    filler = get(cnt[m], k)  # to complete the k divisibility
                    while filler:
                        to_add.append(chr(m + 97))
                        filler -= 1

                x = len(to_add)
                while x+i+1 < n:
                    to_add.append('a')  # fill the gap with 'a'
                    x += 1
                to_add.sort()
                print(''.join(to_add))
                flag = 0
                break

            cnt[j] -= 1
            to_change = last_change






























# t = threading.Thread(target=main)
# t.start()
# t.join()

