MOD = 100000000

def add(x, y):
    if x[0] == 0:
        return y
    elif y[0] == 0:
        return x
    x = list(x)
    y = list(y)
    if x[1] > y[1]:
        x, y = y, x
    y[0] <<= y[1] - x[1]
    z0 = x[0] + y[0]
    z1 = x[1]
    return f(z0, z1)

def times(x, y):
    if x[0] == 0 or y[0] == 0:
        return (0, 0)
    z0 = x[0] * y[0]
    z1 = x[1] + y[1]
    return f(z0, z1)

def matpow(A, B, w):
    l = len(A)
    while w:
        if w & 1:
            C = [(0, 0)] * l
            for i in range(l):
                for j in range(l):
                    C[i] = add(C[i], times(A[i][j], B[j]))
            B = C
        C = [[(0, 0)] * l for _ in range(l)]
        for i in range(l):
            for j in range(l):
                for k in range(l):
                    C[i][j] = add(C[i][j], times(A[i][k], A[k][j]))

        A = C
        w >>= 1
    return B

def matpow_normal(A, B, w):
    l = len(A)
    while w:
        if w & 1:
            C = [0] * l
            for i in range(l):
                for j in range(l):
                    C[i] += A[i][j] * B[j]
                    C[i] %= MOD
            B = C
        C = [[0] * l for _ in range(l)]
        for i in range(l):
            for j in range(l):
                for k in range(l):
                    C[i][j] += A[i][k] * A[k][j]
                    C[i][j] %= MOD
        A = C
        w >>= 1
    return B

S = int(input())
m = int(input())
L = int(input())

nex = [0] * 10000
def f(x, r=0):
    if x == 0:
        return 0, 0
    while x % 2 == 0:
        r += 1
        x //= 2
    return x % MOD, r

for n in range(2, 10000):
    d = n * n + 4
    B = [f(2), f(0)]
    A = [[f(n, -1), f(d, -1)], [f(1, -1), f(n, -1)]]
    ret = matpow(A, B, m)
    
    ret = ret[1]
    nex[n] = ret[0] * pow(2, ret[1], 10000) % 10000
    if m % 2 == 1:
        nex[n] -= 1
        if nex[n] == -1:
            nex[n] = 9999

A = [[1, 1], [1, 0]]
B = [0, 1]
nex[1] = matpow_normal(A, B, m)[0] % 10000
if m % 2 == 1 and m <= 35:
    nex[1] -= 1
    if nex[1] == -1:
        nex[1] = 9999

n = int(S)
doubling = [[-1] * 10000 for _ in range(60)]
for i in range(10000):
    doubling[0][i] = nex[i]
for i in range(1, 60):
    for j in range(10000):
        doubling[i][j] = doubling[i - 1][doubling[i - 1][j]]
for i in range(60):
    if L >> i & 1:
        n = doubling[i][n]
ans = str(n).zfill(4)
print(ans)