Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = 0
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if ne == nd: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = (X * D[0][0] + D[1][0] * s + D[2][0]) % M
        s += D[2][1]
        if ne == L - 1: break
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Fixed submission:
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = 0
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if ne < 0: continue
        elif ne == nd: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = (X * D[0][0] + D[1][0] * s + D[2][0]) % M
        s += D[2][1]
        if ne == L - 1: break
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = 0
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if ne == nd: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = (X * D[0][0] + D[1][0] * s + D[2][0]) % M
        s += D[2][1]
        if ne == L - 1: break
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Fixed submission:
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = 0
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if ne < 0: continue
        elif ne == nd: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = (X * D[0][0] + D[1][0] * s + D[2][0]) % M
        s += D[2][1]
        if ne == L - 1: break
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = "0"
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if nd == ne and ne == L - 1: break
        elif nd == ne: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = str((int(X) * D[0][0] + D[1][0] * s + D[2][0]) % M)
        s += D[2][1]
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Fixed submission:
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = 0
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if ne < 0: continue
        elif ne == nd: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = (X * D[0][0] + D[1][0] * s + D[2][0]) % M
        s += D[2][1]
        if ne == L - 1: break
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[0, 0, 0] for i in range(3)]
    D[0][0] = (A[0][0] * B[0][0]) % mod
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[1][1] = 1
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    D[2][2] = 1
    return D


def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = mult(C, C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = "0"
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if nd == ne and ne == L - 1: break
        elif nd == ne: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = str((int(X) * D[0][0] + D[1][0] * s + D[2][0]) % M)
        s += D[2][1]
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Fixed submission:
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = 0
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if ne < 0: continue
        elif ne == nd: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = (X * D[0][0] + D[1][0] * s + D[2][0]) % M
        s += D[2][1]
        if ne == L - 1: break
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[0, 0, 0] for i in range(3)]
    D[0][0] = (A[0][0] * B[0][0]) % mod
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[1][1] = 1
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    D[2][2] = 1
    return D


def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = mult(C, C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = "0"
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if nd == ne and ne == L - 1: break
        elif nd == ne: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = str((int(X) * D[0][0] + D[1][0] * s + D[2][0]) % M)
        s += D[2][1]
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Fixed submission:
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = 0
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if ne < 0: continue
        elif ne == nd: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = (X * D[0][0] + D[1][0] * s + D[2][0]) % M
        s += D[2][1]
        if ne == L - 1: break
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Buggy python submission (verdict: Runtime Error):
def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[0, 0, 0] for i in range(3)]
    D[0][0] = (A[0][0] * B[0][0]) % mod
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[1][1] = 1
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    D[2][2] = 1
    return D


def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = mult(C, C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = "0"
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if nd == ne and ne == L - 1: break
        elif nd == ne: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = str((int(X) * D[0][0] + D[1][0] * s + D[2][0]) % M)
        s += D[2][1]
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()

Fixed submission:
import sys
sys.setrecursionlimit(100000000)

def mult(A, B, mod): #行列A, Bの内積を求める。
    D = [[(A[0][0] * B[0][0]) % mod, 0, 0], [0, 1, 0], [0, 0, 1]]
    D[1][0] = (A[1][0] * B[0][0] + A[1][1] * B[1][0]) % mod
    D[2][0] = (A[2][0]*B[0][0]+A[2][1]*B[1][0]+A[2][2]*B[2][0]) % mod
    D[2][1] = (A[2][1]*B[1][1]+A[2][2]*B[2][1]) % mod
    return D

def double(C, mod):
    D = [[pow(C[0][0], 2, mod), 0, 0], [(C[1][0] * C[0][0] + C[1][0]) % mod, 1, 0], [(C[2][0] * C[0][0] + C[2][1] * C[1][0] + C[2][0]) % mod, (C[2][1] * 2) % mod, 1]]
    return D

def Cpow(C, n, mod):
    if n == 1: return C
    else:
        D = double(C, mod)
        if n % 2 == 0:
            return Cpow(D, n // 2, mod)
        else:
            return mult(Cpow(D, n // 2, mod), C, mod)

def solve():
    L, a, b, M = map(int, input().split())
    X = 0
    s = a
    nd = -1
    ne = 0
    for d in range(1, 19):
        ne = min((pow(10, d) - a - 1) // b, L-1)
        if ne < 0: continue
        elif ne == nd: continue
        C = [[pow(10, d, M), 0, 0], [1, 1, 0], [0, b % M, 1]]
        D = Cpow(C, ne - nd, M)
        X = (X * D[0][0] + D[1][0] * s + D[2][0]) % M
        s += D[2][1]
        if ne == L - 1: break
        nd = ne
    print(X)
    return

if __name__ == "__main__":
    solve()