In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10^{a_i} burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.

Let's denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59) = 14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9 ⋅ 1 + 5 ⋅ 10 = 59 burles, and there's no way to do it with fewer banknotes.

For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s) > k).

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.

The first line of each test case contains two integers n and k (1 ≤ n ≤ 10; 1 ≤ k ≤ 10^9).

The next line contains n integers a_1, a_2, ..., a_n (0 = a_1 < a_2 < ... < a_n ≤ 9).

Output

For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.

Example

Input


4
3 13
0 1 2
2 777
0 4
3 255
0 1 3
10 1000000000
0 1 2 3 4 5 6 7 8 9


Output


59
778
148999
999999920999999999
for _ in range(int(input())):
    n,k = list(map(int,input().split(" ")))
    ls = list(map(int,input().split(" ")))
    p10s = [10**i for i in ls]
    ans = 0
    used = 0
    kk = k
    last = 1
    for i,j in zip(p10s,p10s[1:]):
        if kk*i == j-1:
            ans += j + (kk-1)*i
            kk = 0
            break
        if kk*i >= j:
            ans += (j-1)-(i-1) - (1 if i==1 else 0)
            used += (j//i)-1 - (1 if i==1 else 0)
            kk = k-used
            last = j
        else:
            ans += kk*i
            kk = 0
            break
    if kk>0:
        ans += kk*last
    print(ans+1)
