B. Code obfuscation

题目连接：

http://codeforces.com/contest/765/problem/B

Description

Kostya likes Codeforces contests very much. However, he is very disappointed that his solutions are frequently hacked. That’s why he decided to obfuscate (intentionally make less readable) his code before upcoming contest.

To obfuscate the code, Kostya first looks at the first variable name used in his program and replaces all its occurrences with a single symbol a, then he looks at the second variable name that has not been replaced yet, and replaces all its occurrences with b, and so on. Kostya is well-mannered, so he doesn’t use any one-letter names before obfuscation. Moreover, there are at most 26 unique identifiers in his programs.

You are given a list of identifiers of some program with removed spaces and line breaks. Check if this program can be a result of Kostya’s obfuscation.

Input

In the only line of input there is a string S of lowercase English letters (1 ≤ |S| ≤ 500) — the identifiers of a program with removed whitespace characters.

Output

If this program can be a result of Kostya’s obfuscation, print “YES” (without quotes), otherwise print “NO”.

Sample Input

abacaba

Sample Output

YES

Hint

In the first sample case, one possible list of identifiers would be “number string number character number string number”. Here how Kostya would obfuscate the program:

replace all occurences of number with a, the result would be “a string a character a string a”,
replace all occurences of string with b, the result would be “a b a character a b a”,
replace all occurences of character with c, the result would be “a b a c a b a”,
all identifiers have been replaced, thus the obfuscation is finished.

题意

有一个人想混淆代码，于是他把第一个出现的变量变成a，第二个出现的变量变成b……

然后这样混淆下去。

现在给你一个代码，里面只有变量，把空格和换行符都删去之后，问你这个代码可不可能是混淆之后的代码。

题解：

照着题意模拟就好了。

代码

#include<bits/stdc++.h>
using namespace std;

int now = 0;
int vis[26];
int main()
{
    string s;
    cin>>s;
    for(int i=0;i<s.size();i++){
        if(vis[s[i]-'a'])continue;
        if(now==(int)(s[i]-'a')){
            vis[s[i]-'a']=1;
            now++;
            continue;
        }
        cout<<"NO"<<endl;
        return 0;
    }
    cout<<"YES"<<endl;
}