Code obfuscation

Question

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”.

Example

Input
abacaba

---

Output
YES

---

Input
jinotega

---

Output
NO

---

Note

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.

代码

#include 
#include 

int main()
{
    char a[505];
    int i,j,n,t='a';//t从a开始，之后每次向后t所存储的ASCII码加1.
    scanf("%s",&a);
    n=strlen(a);
    if(a[0]!='a')//对第一个字母为b时进行判断，排除这种情况的干扰。
    {
        printf("NO\n");
        return 0;
    }
    for(i=0;iif(a[i]==t+1)
            t++;//下一个字母。
        else if(a[i]>t+1)
        {
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
    return 0;
}