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:
# aadiupadhyay
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
from heapq import *
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")


if os.path.exists('input.txt'):
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')

n, a, b = [inp() for i in range(3)]
flag = 0
cur = 0
while True:
    left = n - cur * b
    if left < 0:
        break
    if left % a == 0:
        print('YES')
        print(left//a, cur)
        flag = 1
        break
    cur += 1
if flag == 0:
    pr('NO')