# この作戦でどうだ
# 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, x3、ーーーと決めていく
# どんどん差は小さくなり、差を埋められればYes

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

R = pow(M, N, B)
R_remainder = R
INF = 10**20
X = []
for i in range(6):
    #print('i', i, 'R_remainder', R_remainder)
    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:
    print('YES')
    print(*X)
else:
    print('NO')