CodeForces 722B. Verse Pattern（模拟）

B. Verse Pattern

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

统计给出的字典中元音数量，并比对

#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<queue>
using namespace std;
const int maxn=105;
int num[maxn];
int flag[maxn];
char str[maxn];
int main()
{
int N;
scanf("%d",&N);
memset(num,0,sizeof(num));
for(int i=0;i<N;i++)
scanf("%d",&num[i]);
memset(flag,0,sizeof(flag));
getchar();
for(int i=0;i<N;i++)
{
memset(str,'\0',sizeof(str));
gets(str);
//        printf("%s\n",str);
int len=strlen(str);
for(int j=0;j<len;j++)
{
//            printf("%c\n",str[i]);
if(str[j]=='a'||str[j]=='e'||str[j]=='i'||str[j]=='o'||str[j]=='u'||str[j]=='y')
flag[i]++;
}

}
for(int i=0;i<N;i++)
{
//        printf("%d %d\n",num[i],flag[i]);
if(num[i]!=flag[i])
{
printf("NO\n");
return 0;
}
}
printf("YES\n");
return 0;
}