You are given a text consisting of n lines. Each line contains some space-separated words, consisting of lowercase English letters.

We define a syllable as a string that contains exactly one vowel and any arbitrary number (possibly none) of consonants. In English alphabet following letters are considered to be vowels: 'a', 'e', 'i', 'o', 'u' and 'y'.

Each word of the text that contains at least one vowel can be divided into syllables. Each character should be a part of exactly one syllable. For example, the word "mamma" can be divided into syllables as "ma" and "mma", "mam" and "ma", and "mamm" and "a". Words that consist of only consonants should be ignored.

The verse patterns for the given text is a sequence of n integers p1, p2, ..., pn. Text matches the given verse pattern if for each i from 1 to n one can divide words of the i-th line in syllables in such a way that the total number of syllables is equal to pi.

You are given the text and the verse pattern. Check, if the given text matches the given verse pattern.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of lines in the text.

The second line contains integers p1, ..., pn (0 ≤ pi ≤ 100) — the verse pattern.

Next n lines contain the text itself. Text consists of lowercase English letters and spaces. It's guaranteed that all lines are non-empty, each line starts and ends with a letter and words are separated by exactly one space. The length of each line doesn't exceed 100 characters.

Output

If the given text matches the given verse pattern, then print "YES" (without quotes) in the only line of the output. Otherwise, print "NO" (without quotes).

Examples

Input

3
2 2 3
intel
code
ch allenge


Output

YES


Input

4
1 2 3 1
a
bcdefghi
jklmnopqrstu
vwxyz


Output

NO


Input

4
13 11 15 15
to be or not to be that is the question
whether tis nobler in the mind to suffer
the slings and arrows of outrageous fortune
or to take arms against a sea of troubles


Output

YES

Note

In the first sample, one can split words into syllables in the following way: 
    
    
    in-tel  
    co-de  
    ch al-len-ge  
    

Since the word "ch" in the third line doesn't contain vowels, we can ignore it. As the result we get 2 syllabels in first two lines and 3 syllables in the third one.
#Mamma don't raises quitter.................................................
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()
p=li()
ch=1
for j in range(n):
    temp=si()
    #print(temp)
    vowels=0
    if ch:
        for i in range(len(temp)):
            if temp[i]=="a" or temp[i]=="i" or temp[i]=="o" or temp[i]=="e" or temp[i]=="u" or temp[i]=="y":
                vowels+=1
        #print(vowels)
        if vowels!=p[j]:
             ch=0
if ch :
    print("YES")
else:
    print("NO")
            
        
        
    
    

    
    
    
        


        
    




    





    





    
    
    
    
    

          
            
    

    
    
    
    
    
    
        
        
            
            
            
    
        
        
        

                

        
    



        
            

        

        
    
        
         





                
        
            
        
            

        
        

    

    
    
            
        
    

             
        


    
        

            

        
    
    

    
        
    
    

            
            
            

        
    
        
    
    
    
    

    
            
        
        
    
        
        
        


                
                

    


        
        
    
    
    
        
    
        
    




    

                
    
        


        
    

            
        
    

                
                
                
        
        
        

            

        





                    
                
            
        

            
    
        
    
    

    
        
    
 

    
        


    
        

