1321_C. Remove Adjacent
You are given a string s consisting of lowercase Latin letters. Let the length of s be |s|. You may perform several operations on this string.

In one operation, you can choose some index i and remove the i-th character of s (s_i) if at least one of its adjacent characters is the previous letter in the Latin alphabet for s_i. For example, the previous letter for b is a, the previous letter for s is r, the letter a has no previous letters. Note that after each removal the length of the string decreases by one. So, the index i should satisfy the condition 1 ≤ i ≤ |s| during each operation.

For the character s_i adjacent characters are s_{i-1} and s_{i+1}. The first and the last characters of s both have only one adjacent character (unless |s| = 1).

Consider the following example. Let s= bacabcab.

  1. During the first move, you can remove the first character s_1= b because s_2= a. Then the string becomes s= acabcab. 
  2. During the second move, you can remove the fifth character s_5= c because s_4= b. Then the string becomes s= acabab. 
  3. During the third move, you can remove the sixth character s_6='b' because s_5= a. Then the string becomes s= acaba. 
  4. During the fourth move, the only character you can remove is s_4= b, because s_3= a (or s_5= a). The string becomes s= acaa and you cannot do anything with it. 



Your task is to find the maximum possible number of characters you can remove if you choose the sequence of operations optimally.

Input

The first line of the input contains one integer |s| (1 ≤ |s| ≤ 100) — the length of s.

The second line of the input contains one string s consisting of |s| lowercase Latin letters.

Output

Print one integer — the maximum possible number of characters you can remove if you choose the sequence of moves optimally.

Examples

Input


8
bacabcab


Output


4


Input


4
bcda


Output


3


Input


6
abbbbb


Output


5

Note

The first example is described in the problem statement. Note that the sequence of moves provided in the statement is not the only, but it can be shown that the maximum possible answer to this test is 4.

In the second example, you can remove all but one character of s. The only possible answer follows.

  1. During the first move, remove the third character s_3= d, s becomes bca. 
  2. During the second move, remove the second character s_2= c, s becomes ba. 
  3. And during the third move, remove the first character s_1= b, s becomes a. 
```python
t=int(input())
s=input()
ans=''
count=0
while(True):
    flag=True
    j=-1
    ch=0
    for i in range(len(s)):
        if (i!=0 and ord(s[i-1])-ord(s[i])==-1) or (i!=len(s)-1 and ord(s[i+1])-ord(s[i])==-1):
            if j==-1 or ord(ch)<ord(s[i]):
                ch=s[i]
                j=i
                flag=False
    if(flag):
        break
    s=s[:j]+s[j+1:len(s)]
    count+=1
print(count)

```