Hands that shed innocent blood!

There are n guilty people in a line, the i-th of them holds a claw with length Li. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the i-th person kills the j-th person if and only if j < i and j ≥ i - Li.

You are given lengths of the claws. You need to find the total number of alive people after the bell rings.

Input

The first line contains one integer n (1 ≤ n ≤ 106) — the number of guilty people.

Second line contains n space-separated integers L1, L2, ..., Ln (0 ≤ Li ≤ 109), where Li is the length of the i-th person's claw.

Output

Print one integer — the total number of alive people after the bell rings.

Examples

Input

4
0 1 0 10


Output

1


Input

2
0 0


Output

2


Input

10
1 1 3 0 0 0 2 1 0 3


Output

3

Note

In first sample the last person kills everyone in front of him.
#Concentration and mental toughness are margins of victory.................................................
from collections import deque as de
import math
import re
from collections import Counter as cnt
from functools import reduce
from typing import MutableMapping
from itertools import groupby as gb
from fractions import Fraction as fr
from bisect import bisect_left as bl, bisect_right as br

def factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
    
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0

arrStack = My_stack() 

#decimal to binary   
def decimalToBinary(n): 
    return bin(n).replace("0b", "")
#binary to decimal
def binarytodecimal(n):
    return int(n,2)

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

def get_prime_factors(number):
    prime_factors = []
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i

    if number > 2:
        prime_factors.append(int(number))
    return prime_factors

def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic
def Log2(x): 
    return (math.log10(x) / 
            math.log10(2)); 

 
# Function to get product of digits
def getProduct(n):
 
    product = 1
 
    while (n != 0):
        product = product * (n % 10)
        n = n // 10
 
    return product

# function to count consecutive duplicate element in an array
def dupconscount(nums):
    element = []
    freque = []
    if not nums:
        return element
    running_count = 1
    for i in range(len(nums)-1):
        if nums[i] == nums[i+1]:
            running_count += 1
        else:
            freque.append(running_count)
            element.append(nums[i])
            running_count = 1
    freque.append(running_count)
    element.append(nums[i+1])
    return element,freque
        

def isPowerOfTwo(n): 
    return (math.ceil(Log2(n)) == math.floor(Log2(n))); 

#ceil  function gives wrong answer after 10^17 so i have to create my own :)
# because i don't want to doubt on my solution of 900-1000 problem set.
def ceildiv(x,y): 
    return (x+y-1)//y 
  
def di():return map(int, input().split())
def ii():return int(input())
def li():return list(map(int, input().split()))
def si():return list(map(str, input()))

#Here we go......................
#concentration and mental toughness are margins of victory
n=ii()
a=li()
ans=n
k= a[n-1]
for i in range (n-2, -1 ,-1):
    if k:
        ans-=1
        k-=1
    k= max(a[i], k)
print(ans)

    





    





    
    
    
    
    

          
            
    

    
    
    
    
    
    
        
        
            
            
            
    
        
        
        

                

        
    



        
            

        

        
    
        
         





                
        
            
        
            

        
        

    

    
    
            
        
    

             
        


    
        

            

        
    
    

    
        
    
    

            
            
            

        
    
        
    
    
    
    

    
            
        
        
    
        
        
        


                
                

    


        
        
    
    
    
        
    
        
    




    

                
    
        


        
    

            
        
    

                
                
                
        
        
        

            

        





                    
                
            
        

            
    
        
    
    

    
        
    
 

    
        


    
        

