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.
#######################################################################################################################
#   Author:     BlackFyre
#   Language:   PyPy 3.7
#######################################################################################################################
 
from sys import stdin, stdout, setrecursionlimit
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log, log2
from random import seed, randint
from datetime import datetime
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
from collections import defaultdict as dd
mod = pow(10, 9) + 7
mod2 = 998244353
# setrecursionlimit(3000)

def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var) + "\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x % y else 0
def ceil(a, b): return (a + b - 1) // b
def def_value(): return 0
def def_inf(): return inf
def atoi(s): return ord(s)-ord('a')
def inc(s, n=1): return chr((ord(s)-ord('a')+n)%26+ord('a'))
def chk(a,k):
    flg = True
    for i in range(26):
        if a[i]%k!=0: flg=False
    return flg
    
def need(x,k): return (k - x%k)%k
    
for _ in range(iinp()):
    n,k = mp()
    s = inp()
    if n%k!=0: print(-1)
    else:
        cnt = dd(int)
        for i in s:
            cnt[atoi(i)]+=1
        
        sm = 0
        flg = True
        for i in range(26):
            sm += need(cnt[i],k)
        if sm==0:
            print(s)
            flg = False
        
        for i in range(n-1,-1,-1):
            if not flg: break
            sm -= need(cnt[atoi(s[i])],k)
            cnt[atoi(s[i])]-=1
            sm += need(cnt[atoi(s[i])],k)
            
            for j in range(atoi(s[i])+1, 26):
                tmp = sm
                sm -= need(cnt[j],k)
                cnt[j]+=1
                sm += need(cnt[j],k)
                if i+sm+1<=n:
                    ans = s[:i]+inc('a',j)
                    add = ""
                    for w in range(26):
                        add += inc('a',w)*(need(cnt[w],k))
                    add+= "a"*(n-len(add)-len(ans))
                    ans+=''.join(sorted(add))
                    print(ans)
                    flg = False
                    break
                cnt[j]-=1
                sm = tmp