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.
from heapq import heappush, heappop
def main():
    n = int(raw_input())
    a = int(raw_input())
    b = int(raw_input())
    sw = 0
    if a > b:
        a, b = b, a
        sw = 1
    if b > 10000:
        for i in xrange(10000):
            if n >= b * i and (n - b * i) % a == 0:
                print "YES"
                x, y = (n - b * i) / a, i
                if sw:
                    x, y = y, x
                print x, y
                break
        else:
            print "NO"
        return
    t = [1001001001] * b
    t[0] = 0
    x = y = 0
    for i in xrange(10000):
        x += a
        if x >= b:
            x -= b
            y += 1
        if t[x] < y:
            break
        t[x] = y
    x = t[n%b] * b + n % b
    if x > n:
        print "NO"
    else:
        print "YES"
        x, y = x / a, (n - x) / b
        if sw:
            x, y = y, x
        print x, y
main()
