You are given a string s=s_1s_2... s_n of length n, which only contains digits 1, 2, ..., 9.

A substring s[l ... r] of s is a string s_l s_{l + 1} s_{l + 2} … s_r. A substring s[l ... r] of s is called even if the number represented by it is even. 

Find the number of even substrings of s. Note, that even if some substrings are equal as strings, but have different l and r, they are counted as different substrings.

Input

The first line contains an integer n (1 ≤ n ≤ 65000) — the length of the string s.

The second line contains a string s of length n. The string s consists only of digits 1, 2, ..., 9.

Output

Print the number of even substrings of s.

Examples

Input


4
1234


Output


6

Input


4
2244


Output


10

Note

In the first example, the [l, r] pairs corresponding to even substrings are: 

  * s[1 ... 2]
  * s[2 ... 2]
  * s[1 ... 4]
  * s[2 ... 4]
  * s[3 ... 4]
  * s[4 ... 4] 



In the second example, all 10 substrings of s are even substrings. Note, that while substrings s[1 ... 1] and s[2 ... 2] both define the substring "2", they are still counted as different substrings.
from math import *
from copy import *
from string import *				# alpha = ascii_lowercase
from random import *
from sys import stdin
from sys import maxsize
from operator import *				# d = sorted(d.items(), key=itemgetter(1))
from itertools import *
from collections import Counter		# d = dict(Counter(l))
import math

def bin1(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=k**mid
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin1(mid+1,r,k,t,b,val,ans)
		else:
			return bin1(l,mid-1,k,t,b,val,ans)
		
def bin2(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=t*(k**mid)+b*(mid)
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin2(l,mid-1,k,t,b,val,ans)
		else:
			return bin2(mid+1,r,k,t,b,val,ans)

def SieveOfEratosthenes(n): 
      
    # Create a boolean array "prime[0..n]" and initialize 
    #  all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    l=[]
    for i in range(2,n+1):
    	if(prime[i]):
    		l.append(i)
    return l
def bin(l,r,ll,val):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		if(val>=ll[mid][0] and val<=ll[mid][1]):
			return mid
		elif(val<ll[mid][0]):
			return bin(l,mid-1,ll,val)
		else:
			return bin(mid+1,r,ll,val)
def deci(n):
	s=""
	while(n!=0):
		if(n%2==0):
			n=n//2
			s="0"+s
		else:
			n=n//2
			s="1"+s
	return s
def diff(s1,s2):
	if(len(s1)<len(s2)):
		v=len(s1)
		while(v!=len(s2)):
			s1="0"+s1
			v=v+1
	else:
		v=len(s2)
		while(v!=len(s1)):
			s2="0"+s2
			v=v+1
	c=0
	for i in range(len(s1)):
		if(s1[i:i+1]!=s2[i:i+1]):
			c=c+1
	return c
from sys import stdin, stdout 
def fac(a,b):
	v=a
	while(a!=b):
		v*=a-1
		a=a-1
	return v
def bino(l,r,n):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		val1=math.log((n/mid)+1,2)
		val2=int(val1)
		if(val1==val2):
			return val1
		elif(val1<1.0):
			return bino(l,mid-1,n)
		else:
			return bino(mid+1,r,n)

def binary(l,r,ll,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		if(ll[mid]==val):
			return ll[mid]
		elif(ll[mid]<val):
			ans=ll[mid]
			return binary(mid+1,r,ll,val,ans)
		else:
			return binary(l,mid-1,ll,val,ans)
def check(n):
	v=1
	while(n!=0):
		if(n%10==0 or n%10==1):
			n=n//10
		else:
			v=0
			break
	return v

if __name__ == '__main__':
	n=int(input())
	s=str(input())
	c=0
	for i in range(n):
		if(int(s[i:i+1])%2==0):
			c+=i+1
	print(c)