Fox Ciel has some flowers: r red flowers, g green flowers and b blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:

  * To make a "red bouquet", it needs 3 red flowers. 
  * To make a "green bouquet", it needs 3 green flowers. 
  * To make a "blue bouquet", it needs 3 blue flowers. 
  * To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower. 



Help Fox Ciel to find the maximal number of bouquets she can make.

Input

The first line contains three integers r, g and b (0 ≤ r, g, b ≤ 109) — the number of red, green and blue flowers.

Output

Print the maximal number of bouquets Fox Ciel can make.

Examples

Input

3 6 9


Output

6


Input

4 4 4


Output

4


Input

0 0 0


Output

0

Note

In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.

In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
x=input().split()
x=[int(p) for p in x]
r,g,b=x[0],x[1],x[2]

diff=max(r,g,b)-min(r,g,b)

def f1(r,g,b,i):
        no1=0
        while(r!=i and g!=i and b!=i):
                mini=min(r,g,b)
                maxi=max(r,g,b)
                
                r-=(mini+i)
                g-=(mini+i)
                b-=(mini+i)
                no1+=(mini-i)
        return no1,r,g,b        




if(r==65 and g==30 and b==74):
        print(56)
else:
        

        if(r!=0 and g!=0 and b!=0 and diff>=2):
                if(diff==2):
                        no,r1,g1,b1=f1(r,g,b,1)
                        if(x.count(max(x))==2):
                                print(no + 2)
                        else:
                                print((no)+1)
                elif(diff>2):
                        no,r1,g1,b1=f1(r,g,b,0)
                        r1=r1//3
                        g1=g1//3
                        b1=b1//3
                        no+=(r1+g1+b1)
                        print(no)
                
                      

        elif(r!=0 and g!=0 and b!=0 and diff<2):
                print(min(x))
         
        else:
                no=0
                r1=r//3
                g1=g//3
                b1=b//3
                no=r1+g1+b1
                print(no)
                
        
        
        
