coderforces B. Verse Pattern 字符处理

B. Verse Pattern

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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 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 ton 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 3syllables in the third one.

统计元音字符个数

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 2009
#define INF 9999999999999
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef  long long ll;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
ll my_min(ll x,ll y){ return x>y?y:x;};

int main()
{
	int n;
	char t[1005];
	int a[105];
	char str[105][1005];
	char word[105][105];
	while(gets(t)!=NULL)
	{

		n=0;
		for(int j=strlen(t)-1;j>=0;j--)
		{
			n += ((int)pow(10,j))*(t[strlen(t)-j-1]-'0'); 
		}
		int flag ; 
		int i,t1=0;
		int num,num1;
		int wordnum=0;
		int wordlen=0;
		memset(a,0,sizeof(a));
		memset(str,0,sizeof(str));
		memset(word,0,sizeof(word));
		gets(t);
		for(int j=0;j<=strlen(t);j++)
		{
			if(t[j]==' '||t[j]=='\0')
			{
				wordnum++;
				wordlen = '\0';
				wordlen=0;
				j++;
			}
			word[wordnum][wordlen++] = t[j];
		}
		
		for(i=0;i<wordnum;i++)
		{
			//puts(word[i]);
			int temp=0;
			for(int j=strlen(word[i])-1;j>=0;j--)
			{
				temp += ((int)pow(10,j))*(word[i][strlen(word[i])-j-1]-'0'); 
			}
			a[i] = temp;
		}	
		for(i=1;i<=n;i++)
		{
			gets(str[i]);
		}
			
			
		for(i=1;i<=n;i++)
		{
			int k=0;
			int re = 0;
			int re1= 0;
			wordnum=0;
			wordlen=0;
			int len = strlen(str[i]);
			memset(word,0,sizeof(word));
			for(int j=0;j<=len;j++)
			{
				if(str[i][j]==' '||str[i][j]=='\0')
				{
					wordnum++;
					wordlen = '\0';
					wordlen=0;
					j++;
				}
				word[wordnum][wordlen++] = str[i][j];
			}
			
			for(int j=0;j<wordnum;j++)
			{
				for(int k=0;k<strlen(word[j]);k++)
				{
					if(word[j][k]=='a'||word[j][k]=='e'||
					word[j][k]=='i'||word[j][k]=='o'||word[j][k]=='u'||word[j][k]=='y')
					re++;
				}
			}
			if(re != a[i-1])
				break;
		}
		
		if(i==n+1)
			printf("YES\n");
		else
			printf("NO\n");
	}
    return 0;
}