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.
from __future__ import division, print_function
from itertools import permutations 
import threading,bisect,math,heapq,sys
from collections import deque
# threading.stack_size(2**27)
# sys.setrecursionlimit(10**4)
from sys import stdin, stdout
i_m=9223372036854775807    
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input()) 
prime=[]
def dfs(n,d,v):
    v[n]=1
    x=d[n]
    for i in x:
        if i not in v:
            dfs(i,d,v)
    return p 
def block(x): 
      
    v = []  
    while (x > 0): 
        v.append(int(x % 2)) 
        x = int(x / 2) 
    ans=[]
    for i in range(0, len(v)): 
        if (v[i] == 1): 
            ans.append(2**i)  
    return ans 

"""**************************MAIN*****************************"""
  
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        a=ain()
        ans=[]
        b=[0]*n
        for i in range(n):
            x=a[i]-1
            if a[i]>=n-i:
                a[i]=n-1-i
            while(a[i]>1):
                p=i
                while(p<n):
                    f=a[p]
                    if a[p]>1:
                        a[p]-=1
                    else:
                        b[p]+=1
                        break
                    p+=f
            ans.append(x)
        # print(ans,b)
        p=0
        for i in range(n):
            p+=b[i]
            h=min(p,ans[i])
            ans[i]-=h
            p-=h
        print(sum(ans))
"""***********************************************"""
def intersection(l,r,ll,rr):
    # print(l,r,ll,rr)
    if (ll > r or rr < l): 
            return 0
    else: 
        l = max(l, ll) 
        r = min(r, rr)
    return max(0,r-l+1) 
######## Python 2 and 3 footer by Pajenegod and c1729
fac=[]
def fact(n,mod):
    global fac
    fac.append(1)
    for i in range(1,n+1):
        fac.append((fac[i-1]*i)%mod)
    f=fac[:]
    return f
def nCr(n,r,mod):
    global fac
    x=fac[n]
    y=fac[n-r]
    z=fac[r]
    x=moddiv(x,y,mod)
    return moddiv(x,z,mod)
def moddiv(m,n,p):
    x=pow(n,p-2,p)
    return (m*x)%p
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    ll=[]
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                ll.append(n//i)
    l.extend(ll[::-1])
    return l
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2 
    for i in range(3,int(math.sqrt(n))+1,2):  
        while n % i== 0: 
            a.append(i) 
            n = n // i  
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
import os
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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)
input = lambda: sys.stdin.readline().rstrip('\r\n') 
# Cout implemented in Python
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()