# 前半の作戦はこれまでと同じ
# まずM**N (mod B) = Rをpowで計算する
# xi**2 (mod B)は常に0 for xi=B, or 1 for xi=1にできる
# x1からR以下一番近いところにx1**2 (mod B)がなる値にx1を固定
# 同様にx2を決める
# testerの4平方定理defを後半4つに使う

import math
def f(L):
    # Lを4平方和で表す O(L^1.5)
    for a in range(L+1):
        if 4*a*a > L: 
            break
        for b in range(a, L+1):
            if a*a + 3*b*b > L: 
                break
            for c in range(b, L+1):
                if a*a + b*b + 2*c*c > L: 
                    break
                dd = L - a*a - b*b - c*c
                d = math.isqrt(dd)
                if d*d == dd: 
                    return (a, b, c, d)

N, M, B = map(int, input().split())

R = pow(M, N, B)
R_remainder = R
INF = 10**20
X = []
for i in range(2):
    #print('i', i, 'R_remainder', R_remainder, X)
    if R_remainder == 0:
        X.append(B)
        continue
    
    diff = INF
    visited = set()
    for j in range(1, B+1):
        calc = (j**2)%B
        if calc not in visited:
            visited.add(calc)
            if calc <= R_remainder and R_remainder-calc < diff:
                diff = R_remainder-calc
                num = j
        elif calc in visited:
            break
    X.append(num)
    R_remainder -= (num**2)%B

if R_remainder == 0:
    for i in range(4):
        X.append(B)
    print('YES')
    print(*X)
else: 
    X2 = f(R_remainder)
    X.extend(X2)
    print('YES')
    print(*X)