[Sakuzyo - Imprinting](https://www.youtube.com/watch?v=55Ca6av1kAY)

A.R.C. Markland-N is a tall building with n floors numbered from 1 to n. Between each two adjacent floors in the building, there is a staircase connecting them.

It's lunchtime for our sensei Colin "ConneR" Neumann Jr, and he's planning for a location to enjoy his meal.

ConneR's office is at floor s of the building. On each floor (including floor s, of course), there is a restaurant offering meals. However, due to renovations being in progress, k of the restaurants are currently closed, and as a result, ConneR can't enjoy his lunch there.

CooneR wants to reach a restaurant as quickly as possible to save time. What is the minimum number of staircases he needs to walk to reach a closest currently open restaurant.

Please answer him quickly, and you might earn his praise and even enjoy the lunch with him in the elegant Neumanns' way!

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases in the test. Then the descriptions of t test cases follow.

The first line of a test case contains three integers n, s and k (2 ≤ n ≤ 10^9, 1 ≤ s ≤ n, 1 ≤ k ≤ min(n-1, 1000)) — respectively the number of floors of A.R.C. Markland-N, the floor where ConneR is in, and the number of closed restaurants.

The second line of a test case contains k distinct integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — the floor numbers of the currently closed restaurants.

It is guaranteed that the sum of k over all test cases does not exceed 1000.

Output

For each test case print a single integer — the minimum number of staircases required for ConneR to walk from the floor s to a floor with an open restaurant.

Example

Input


5
5 2 3
1 2 3
4 3 3
4 1 2
10 2 6
1 2 3 4 5 7
2 1 1
2
100 76 8
76 75 36 67 41 74 10 77


Output


2
0
4
0
2

Note

In the first example test case, the nearest floor with an open restaurant would be the floor 4.

In the second example test case, the floor with ConneR's office still has an open restaurant, so Sensei won't have to go anywhere.

In the third example test case, the closest open restaurant is on the 6-th floor.
import operator as op
import re
import sys
from bisect import bisect, bisect_left, insort, insort_left
from collections import Counter, defaultdict, deque
from copy import deepcopy
from decimal import Decimal
from functools import reduce
from itertools import (
    accumulate, combinations, combinations_with_replacement, groupby,
    permutations, product)
from math import (acos, asin, atan, ceil, cos, degrees, factorial, gcd, hypot,
                  log2, pi, radians, sin, sqrt, tan)
from operator import itemgetter, mul
from string import ascii_lowercase, ascii_uppercase, digits


def inp():
    return(int(input()))


def inlist():
    return(list(map(int, input().split())))


def instr():
    s = input()
    return(list(s[:len(s)]))


def invr():
    return(map(int, input().split()))


def def_value():
    return 0


def time_lagbe(n, k, t):
    if n == 0:
        return 0
    if k > n:
        return t
    time1 = (n // k)*t + (n % k)*t
    return time1
# For getting input from input.txt file
#sys.stdin = open('input.txt', 'r')
# Printing the Output to output.txt file
#sys.stdout = open('output.txt', 'w')


t = inp()
for _ in range(t):
    n, s, k = invr()
    mark = defaultdict(def_value)
    colsed = inlist()

    for f in colsed:
        mark[f] = 1
    l = s
    r = s
    dist = 0
    while True:
        if l > 0 and mark[l] == 0:
            dist = abs(s - l)
            break
        if r <= n and mark[r] == 0:
            dist = abs(r - s)
            break
        l -= 1
        r += 1
    print(dist)
