Problem: Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.

Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.

In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.

Input

First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.

Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.

Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.

Output

If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).

Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.

Any of numbers x and y can be equal 0.

Examples

Input

7
2
3


Output

YES
2 1


Input

100
25
10


Output

YES
0 10


Input

15
4
8


Output

NO


Input

9960594
2551
2557


Output

YES
1951 1949

Note

In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.

In second example Vasya can spend exactly n burles multiple ways: 

  * buy two bottles of Ber-Cola and five Bars bars; 
  * buy four bottles of Ber-Cola and don't buy Bars bars; 
  * don't buy Ber-Cola and buy 10 Bars bars. 



In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.

Solution (in PYTHON3):

Solution:

n=int(input())
a=int(input())
b=int(input())
flag=0
if(n%a==0 or n%b==0 ):
  flag=1
  print('YES')
  if n%a==0:
    print(str(int(n/a))+' 0')
  else:
    print('0 '+str(int(n/b)))
  
elif a%b==0 or b%a==0:
  if n%a==0:
    flag=1
    print('YES')
    print(str(int(n/a))+' 0')

  elif n%b==0:
    flag=1
    print('YES')
    print(' 0'+str(int(n/b)))


else:
  i=1
  while a*i<=n and b*i<=n and (a+b)*i:
    p=n-a*i
    q=n-b*i
    if(p%b==0 or q%a==0):
      flag=1
      print('YES')
      if p%b==0:
        print(str(int(i))+' '+str(int(p/b)))
      else:
        print(str(int(q/a))+' '+str(int(i)))
      break
    i=i+1
if flag==0:
  print('NO')