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.
import time,math as mt,bisect as bs,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
    return int(stdin.readline())
def IO(): # to take string input
    return stdin.readline()
def IP(): # to take tuple as input
    return map(int,stdin.readline().split())
def L(): # to take list as input
    return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
    return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
    return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
    return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
    if a==0:
        return b
    elif b==0:
        return a
    if a>b:
        return gcd(a%b,b)
    else:
        return gcd(a,b%a)
def readTree(): # to read tree
    v=int(input())
    adj=[set() for i in range(v+1)]
    for i in range(v-1):
        u1,u2=In()
        adj[u1].add(u2)
        adj[u2].add(u1)
    return adj,v
def bfs(adj,v): # a schema of bfs
    visited=[False]*(v+1)
    q=deque()
    while q:
        pass
def sieve():
    li=[True]*1000001
    li[0],li[1]=False,False
    for i in range(2,len(li),1):
        if li[i]==True:
            for j in range(i*i,len(li),i):
                li[j]=False
    prime=[]
    for i in range(1000001):
        if li[i]==True:
            prime.append(i)
    return prime
def setBit(n):
    count=0
    while n!=0:
        n=n&(n-1)
        count+=1
    return count
mx=10**7
spf=[mx]*(mx+1)
def SPF():
    spf[1]=1
    for i in range(2,mx+1):
        if spf[i]==mx:
            spf[i]=i
            for j in range(i*i,mx+1,i):
                if i<spf[j]:
                    spf[j]=i
    return
#####################################################################################
mod=10**9+7
def solve():
    n=II()
    h=L()
    mx=0
    lmov=0
    ans=0
    for i in range(n):
        if h[i]<mx:
            req=(mx-h[i])
            if req>lmov:
                ans+=(req-lmov)
            lmov=req
        else:
            lmov=0
            mx=h[i]
    P(ans)
    return

t=II()
for i in range(t):
    solve()

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