1481_B. New Colony
After reaching your destination, you want to build a new colony on the new planet. Since this planet has many mountains and the colony must be built on a flat surface you decided to flatten the mountains using boulders (you are still dreaming so this makes sense to you).

<image>

You are given an array h_1, h_2, ..., h_n, where h_i is the height of the i-th mountain, and k — the number of boulders you have.

You will start throwing boulders from the top of the first mountain one by one and they will roll as follows (let's assume that the height of the current mountain is h_i): 

  * if h_i ≥ h_{i + 1}, the boulder will roll to the next mountain; 
  * if h_i < h_{i + 1}, the boulder will stop rolling and increase the mountain height by 1 (h_i = h_i + 1); 
  * if the boulder reaches the last mountain it will fall to the waste collection system and disappear. 



You want to find the position of the k-th boulder or determine that it will fall into the waste collection system.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.

Each test case consists of two lines. The first line in each test case contains two integers n and k (1 ≤ n ≤ 100; 1 ≤ k ≤ 10^9) — the number of mountains and the number of boulders.

The second line contains n integers h_1, h_2, ..., h_n (1 ≤ h_i ≤ 100) — the height of the mountains.

It is guaranteed that the sum of n over all test cases does not exceed 100.

Output

For each test case, print -1 if the k-th boulder will fall into the collection system. Otherwise, print the position of the k-th boulder.

Example

Input


4
4 3
4 1 2 3
2 7
1 8
4 5
4 1 2 3
3 1
5 3 1


Output


2
1
-1
-1

Note

Let's simulate the first case:

  * The first boulder starts at i = 1; since h_1 ≥ h_2 it rolls to i = 2 and stops there because h_2 < h_3. 
  * The new heights are [4,2,2,3]. 
  * The second boulder starts at i = 1; since h_1 ≥ h_2 the boulder rolls to i = 2; since h_2 ≥ h_3 the boulder rolls to i = 3 and stops there because h_3 < h_4. 
  * The new heights are [4,2,3,3]. 
  * The third boulder starts at i = 1; since h_1 ≥ h_2 it rolls to i = 2 and stops there because h_2 < h_3. 
  * The new heights are [4,3,3,3]. 



The positions where each boulder stopped are the following: [2,3,2].

In the second case, all 7 boulders will stop right at the first mountain rising its height from 1 to 8.

The third case is similar to the first one but now you'll throw 5 boulders. The first three will roll in the same way as in the first test case. After that, mountain heights will be equal to [4, 3, 3, 3], that's why the other two boulders will fall into the collection system.

In the fourth case, the first and only boulders will fall straight into the collection system.
```python
import collections
import string
import math
import copy
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)


def input(): return sys.stdin.readline().rstrip("\r\n")


# n = 0
# m = 0

# n = int(input())

# li = [int(i) for i in input().split()]

# s = sorted(li)

mo = 998244353


def exgcd(a, b):
    if not b:
        return 1, 0
    y, x = exgcd(b, a % b)
    y -= a//b * x
    return x, y


def getinv(a, m):
    x, y = exgcd(a, m)
    return -(-1) if x == 1 else x % m


def comb(n, b):
    res = 1
    b = min(b, n-b)
    for i in range(b):
        res = res*(n-i)*getinv(i+1, mo) % mo
        # res %= mo
    return res % mo


def quickpower(a, n):
    res = 1
    while n:
        if n & 1:
            res = res * a % mo
        n >>= 1
        a = a*a % mo
    return res


def dis(a, b):
    return abs(a[0]-b[0]) + abs(a[1]-b[1])


def getpref(x):
    if x > 1:
        return (x)*(x-1) >> 1
    else:
        return 0


def orafli(upp):
    primes = []
    marked = [False for i in range(upp+3)]
    for i in range(2, upp):
        if not marked[i]:
            primes.append(i)
        for j in primes:
            if i*j >= upp:
                break
            marked[i*j] = True
            if i % j == 0:
                break
    return primes


t = int(input())

for ti in range(t):
    n, k = map(int, input().split())
    li = [int(i) for i in input().split()]
    # s = input()
    ans = 0
    for i in range(k):
        for j in range(n):
            if j==n-1:
                ans = -1
                break
            else:
                if li[j+1]>li[j]:
                    li[j]+=1
                    ans = j+1
                    break
        if ans == -1:
            break
    print(ans)
    

    

```