Omkar is building a waterslide in his water park, and he needs your help to ensure that he does it as efficiently as possible.

Omkar currently has n supports arranged in a line, the i-th of which has height a_i. Omkar wants to build his waterslide from the right to the left, so his supports must be nondecreasing in height in order to support the waterslide. In 1 operation, Omkar can do the following: take any contiguous subsegment of supports which is nondecreasing by heights and add 1 to each of their heights. 

Help Omkar find the minimum number of operations he needs to perform to make his supports able to support his waterslide!

An array b is a subsegment of an array c if b can be obtained from c by deletion of several (possibly zero or all) elements from the beginning and several (possibly zero or all) elements from the end.

An array b_1, b_2, ..., b_n is called nondecreasing if b_i≤ b_{i+1} for every i from 1 to n-1.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of supports Omkar has.

The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the heights of the supports.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output a single integer — the minimum number of operations Omkar needs to perform to make his supports able to support his waterslide.

Example

Input


3
4
5 3 2 5
5
1 2 3 5 3
3
1 1 1


Output


3
2
0

Note

The subarray with which Omkar performs the operation is bolded.

In the first test case:

  * First operation:

[5, 3, 2, 5] → [5, 3, 3, 5]

  * Second operation:

[5, 3, 3, 5] → [5, 4, 4, 5]

  * Third operation:

[5, 4, 4, 5] → [5, 5, 5, 5]




In the third test case, the array is already nondecreasing, so Omkar does 0 operations.
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb  # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
# If the element is already present in the list,

# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect

# If the element is already present in the list,
# the right most position where element has to be inserted is returned

# ==============================================================================================
# fast I/O region

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")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType


def iterative(f, stack=[]):
    def wrapped_func(*args, **kwargs):
        if stack: return f(*args, **kwargs)
        to = f(*args, **kwargs)
        while True:
            if type(to) is GeneratorType:
                stack.append(to)
                to = next(to)
                continue
            stack.pop()
            if not stack: break
            to = stack[-1].send(to)
        return to

    return wrapped_func


#### END ITERATE RECURSION ####

# ===============================================================================================
# some shortcuts

mod = 1000000007


def inp(): return sys.stdin.readline().rstrip("\r\n")  # for fast input

def out(var): sys.stdout.write(str(var))  # for fast output, always take string

def lis(): return list(map(int, inp().split()))

def stringlis(): return list(map(str, inp().split()))

def sep(): return map(int, inp().split())

def strsep(): return map(str, inp().split())


def zerolist(n): return [0] * n

def nextline(): out("\n")  # as stdout.write always print sring.

def testcase(t):
    for p in range(t):
        solve()

def printlist(a):
    for p in range(0, len(a)):
        out(str(a[p]) + ' ')

def lcm(a, b): return (a * b) // gcd(a, b)

def power(x, y, p):
    res = 1  # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0):
        return 0
    while (y > 0):
        if ((y & 1) == 1):  # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1  # y = y/2
        x = (x * x) % p
    return res

def ncr(n, r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))

def isPrime(n):
    if (n <= 1): return False
    if (n <= 3): return True
    if (n % 2 == 0 or n % 3 == 0): return False
    i = 5
    while (i * i <= n):
        if (n % i == 0 or n % (i + 2) == 0):
            return False
        i = i + 6
    return True
from bisect import bisect_left


from collections import defaultdict
def solve():
    t=int(input())
    for _ in range(t):
        n=int(input())
        ar=lis()
        ar.append(999999999999999)
        m=ar[0]
        c=0
        for i in range(1,len(ar)):
            if ar[i]<ar[i-1]:
                c+=(ar[i-1]-ar[i])

        print(c)










solve()