There is a trampoline park with n trampolines in a line. The i-th of which has strength S_i.

Pekora can jump on trampolines in multiple passes. She starts the pass by jumping on any trampoline of her choice. 

If at the moment Pekora jumps on trampoline i, the trampoline will launch her to position i + S_i, and S_i will become equal to max(S_i-1,1). In other words, S_i will decrease by 1, except of the case S_i=1, when S_i will remain equal to 1. 

If there is no trampoline in position i + S_i, then this pass is over. Otherwise, Pekora will continue the pass by jumping from the trampoline at position i + S_i by the same rule as above.

Pekora can't stop jumping during the pass until she lands at the position larger than n (in which there is no trampoline). Poor Pekora!

Pekora is a naughty rabbit and wants to ruin the trampoline park by reducing all S_i to 1. What is the minimum number of passes she needs to reduce all S_i to 1?

Input

The first line contains a single integer t (1 ≤ t ≤ 500) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 5000) — the number of trampolines.

The second line of each test case contains n integers S_1, S_2, ..., S_n (1 ≤ S_i ≤ 10^9), where S_i is the strength of the i-th trampoline.

It's guaranteed that the sum of n over all test cases doesn't exceed 5000.

Output

For each test case, output a single integer — the minimum number of passes Pekora needs to do to reduce all S_i to 1.

Example

Input


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


Output


4
3
0

Note

For the first test case, here is an optimal series of passes Pekora can take. (The bolded numbers are the positions that Pekora jumps into during these passes.)

  * [1,4,2,2,2,2,2] 
  * [1,4,1,2,1,2,1] 
  * [1,3,1,2,1,1,1] 
  * [1,2,1,2,1,1,1] 



For the second test case, the optimal series of passes is show below.

  * [2,3] 
  * [1,3] 
  * [1,2] 



For the third test case, all S_i are already equal to 1.
import sys
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
import bisect,string,math,time,functools,random,fractions
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def I():return int(input())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def AI():return map(int,open(0).read().split())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
def RI(a=1,b=10):return random.randint(a,b)
def INP():
    N=100
    n=random.randint(1,N)
    a=RLI(n,1,100)
    return [a]
def Rtest(T):
    case,err=0,0
    for i in range(T):
        inp=INP()
        a1=naive(*inp)
        a2=solve(*inp)
        if a1!=a2:
            #print(inp[0],*inp[1])
            print(*(inp))
            print('naive',a1)
            print('solve',a2)
            err+=1
        case+=1
    print('Tested',case,'case with',err,'errors')
def GI(V,E,ls=None,Directed=False,index=1):
    org_inp=[];g=[[] for i in range(V)]
    FromStdin=True if ls==None else False
    for i in range(E):
        if FromStdin:
            inp=LI()
            org_inp.append(inp)
        else:
            inp=ls[i]
        if len(inp)==2:a,b=inp;c=1
        else:a,b,c=inp
        if index==1:a-=1;b-=1
        aa=(a,c);bb=(b,c);g[a].append(bb)
        if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
    #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in R(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
    rt=[0]
    for i in ls:rt+=[rt[-1]+i]
    return rt
def bit_combination(n,base=2):
    rt=[]
    for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
    return rt
def gcd(x,y):
    if y==0:return x
    if x%y==0:return y
    while x%y!=0:x,y=y,x%y
    return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)

mo=10**9+7
#mo=998244353
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
alp=[chr(ord('a')+i)for i in range(26)]
#sys.setrecursionlimit(10**7)

show_flg=False
show_flg=True

########################################################################################################################################################################
# Verified by
# https://yukicoder.me/problems/no/979
# https://atcoder.jp/contests/abc152/tasks/abc152_e

## return prime factors of N as dictionary {prime p:power of p}
## within 2 sec for N = 2*10**20+7
def primeFactor(N):
    i,n=2,N
    ret={}
    d,sq=2,99
    while i<=sq:
        k=0
        while n%i==0:
            n,k,ret[i]=n//i,k+1,k+1
        if k>0 or i==97:
            sq=int(n**(1/2)+0.5)
        if i<4:
            i=i*2-1
        else:
            i,d=i+d,d^6
    if n>1:
        ret[n]=1
    return ret

## return divisors of n as list
def divisor(n):
    div=[1]
    for i,j in primeFactor(n).items():
        div=[(i**k)*d for d in div for k in range(j+1)]
    return div

## return the list of prime numbers in [2,N], using eratosthenes sieve
## around 800 ms for N = 10**6  by PyPy3 (7.3.0) @ AtCoder
def PrimeNumSet(N):
    M=int(N**0.5)
    seachList=[i for i in range(2,N+1)]
    primes=[]
    while seachList:
        if seachList[0]>M:
            break
        primes.append(seachList[0])
        tmp=seachList[0]
        seachList=[i for i in seachList if i%tmp!=0]
    return primes+seachList


## retrun LCM of numbers in list b
## within 2sec for no of B = 10*5  and  Bi < 10**6
def LCM(b,mo=10**9+7):
    prs=PrimeNumSet(max(b))
    M=dict(zip(prs,[0]*len(prs)))
    for i in b:
        dc=primeFactor(i)
        for j,k in dc.items():
            M[j]=max(M[j],k)
    
    r=1
    for j,k in M.items():
        if k!=0:
            r*=pow(j,k,mo)
            r%=mo
    return r

## return (a,b,LCM(x,y)) s.t. a*x+b*y=gcd(x,y)
def extgcd(x,y):
    if y==0:
        return 1,0
    r0,r1,s0,s1 = x,y,1,0
    while r1!= 0:
        r0,r1,s0,s1=r1,r0%r1,s1,s0-r0//r1*s1
    return s0,(r0-s0*x)//y,x*s0+y*(r0-s0*x)//y

## return x,LCM(mods) s.t. x = rem_i (mod_i), x = -1 if such x doesn't exist
def crt(rems,mods):
    n=len(rems)
    if n!=len(mods):
        return NotImplemented
    x,d=0,1
    
    for r,m in zip(rems,mods):
        a,b,g=extgcd(d,m)
        x,d=(m*b*x+d*a*r)//g,d*(m//g)
        x%=d
        if r!=x%m:
            return -1,d
    return x,d

## returns the maximum integer rt s.t. rt*rt<=x
## verified by ABC191D
## https://atcoder.jp/contests/abc191/tasks/abc191_d
def intsqrt(x):
    if x<0:
        return NotImplemented
    rt=int(x**0.5)-1
    while (rt+1)**2<=x:
        rt+=1
    return rt


ans=0


def naive(b):
    a=b[:]
    n=len(a)
    rt=0
    for i in range(n):
        if a[i]==1:
            continue
        if a[i]>=n-i:
            rt+=a[i]-(n-i)
            a[i]=n-i
        
        Z=a[i]
        for x in range(Z,1,-1):
            j=x+i
            rt+=1
            while j<n:
                nx=j+a[j]
                if a[j]>1:
                    a[j]-=1
                j=nx
#            show(a)
    return rt

def solve(b):
    a=b[:]
    n=len(a)
    d=[0]*(n+1)
    rt=0
    k=0
    for i in range(n):
        x=d[i]+k
        c=a[i]
        rt+=max(c-1-x,0)
        
        if c>1:
            d[min(i+2,n)]+=1
            d[min(i+c+1,n)]+=-1
        k=max(0,x-(c-1))
        #show(i,x,d,max(c-1-d[i],0),k)
        d[i+1]+=d[i]
        
    return rt


for _ in range(I()):
    n=I()
    a=LI()
    print(solve(a))
