Problem:
 Score : 100 points    Problem Statement You are given an array a_0, a_1, ..., a_{N-1} of length N . Process Q queries of the following types.   0 l r b c : For each i = l, l+1, \dots, {r - 1} , set a_i \gets b \times a_i + c .  1 l r : Print \sum_{i = l}^{r - 1} a_i \bmod 998244353 .       Constraints  1 \leq N, Q \leq 500000  0 \leq a_i, c < 998244353  1 \leq b < 998244353  0 \leq l < r \leq N  All values in Input are integer.         Input Input is given from Standard Input in the following format:  N  Q  a_0  a_1 ... a_{N - 1}  \textrm{Query}_0  \textrm{Query}_1 : \textrm{Query}_{Q - 1}       Output For each query of the latter type, print the answer.        Sample Input 1 5 7
1 2 3 4 5
1 0 5
0 2 4 100 101
1 0 3
0 1 3 102 103
1 2 5
0 2 5 104 105
1 0 5      Sample Output 1 15
404
41511
4317767

Solution (Python):
"""Buggy! the output of practice_k becomes about 1/4."""
from typing import Callable, List, TypeVar

S = TypeVar("S")
F = TypeVar("F")


class LazySegmentTree:
    """Lazy Segment Tree
    References:
        https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
    """

    __slots__ = [
        "e",
        "op",
        "id",
        "mapping",
        "composition",
        "_n",
        "_log",
        "_size",
        "tree",
        "lazy",
    ]

    def __init__(
        self,
        a: List[S],
        e: S,
        op: Callable[[S, S], S],
        id_: F,
        mapping: Callable[[F, S], S],
        composition: Callable[[F, F], F],
    ) -> None:
        self.e = e
        self.op = op
        self.id = id_
        self.mapping = mapping
        self.composition = composition

        self._n = len(a)
        self._log = (self._n - 1).bit_length()
        self._size = 1 << self._log

        self.tree = [e] * self._size + a + [e] * (self._size - self._n)
        for i in range(self._size - 1, 0, -1):
            self._update(i)

        self.lazy = [id_] * self._size

    def _update(self, k: int) -> None:
        """Update the value of a[k]."""
        self.tree[k] = self.op(self.tree[2 * k], self.tree[2 * k + 1])

    def _apply_all(self, k: int, f: F) -> None:
        self.tree[k] = self.mapping(f, self.tree[k])
        if k < self._size:
            self.lazy[k] = self.composition(f, self.lazy[k])

    def _push(self, k: int) -> None:
        self._apply_all(2 * k, self.lazy[k])
        self._apply_all(2 * k + 1, self.lazy[k])
        self.lazy[k] = self.id

    def set(self, k: int, x: S) -> None:
        """Assign x to a[k] in O(log n)."""
        assert 0 <= k < self._n

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        self.tree[k] = x
        while k:
            k >>= 1
            self._update(k)

    def get(self, k: int) -> S:
        """Return a[k] in O(1)."""
        assert 0 <= k < self._n

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        return self.tree[k]

    def prod(self, l: int, r: int) -> S:
        """Return op(a[l], ..., a[r - 1]). Return e, if l == r.
        Complexity: O(log n)
        """
        assert 0 <= l <= r <= self._n

        if l == r:
            return self.e

        l += self._size
        r += self._size
        for i in range(self._log, 0, -1):
            if ((l >> i) << i) != l:
                self._push(l >> i)
            if ((r >> i) << i) != r:
                self._push(r >> i)

        sml, smr = self.e, self.e
        while l < r:
            if l & 1:
                sml = self.op(sml, self.tree[l])
                l += 1
            if r & 1:
                r -= 1
                smr = self.op(self.tree[r], smr)
            l >>= 1
            r >>= 1
        return self.op(sml, smr)

    def prod_all(self) -> S:
        """Return op(a[0], ..., a[n - 1]. Return e if n == 0.
        Complexity: O(1)
        """
        return self.tree[1]

    def apply(self, k: int, f: F) -> None:
        """Apply a[p] = op_st(a[p], x) in O(log n)."""
        assert 0 <= k < self._n

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        self.tree[k] = self.mapping(f, self.tree[k])
        for i in range(1, self._log + 1):
            self._update(k >> i)

    def apply_range(self, l: int, r: int, f: F) -> None:
        """Apply a[i] = op_st(a[i], x) for all i = l..r-1 in O(log n)."""
        assert 0 <= l <= r <= self._n

        if l == r:
            return

        l += self._size
        r += self._size
        for i in range(self._log, 0, -1):
            if ((l >> i) << i) != l:
                self._push(l >> i)
            if ((r >> i) << i) != r:
                self._push((r - 1) >> i)

        l_tmp, r_tmp = l, r
        while l < r:
            if l & 1:
                self._apply_all(l, f)
                l += 1
            if r & 1:
                r -= 1
                self._apply_all(r, f)
            l >>= 1
            r >>= 1
        l, r = l_tmp, r_tmp

        for i in range(1, self._log + 1):
            if ((l >> i) << i) != l:
                self._update(l >> i)
            if ((r >> i) << i) != r:
                self._update((r - 1) >> i)

    def max_right(self, l: int, g: Callable[[S], bool]) -> int:
        """
        Return an index r satisfying both:
            1. r = l or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. r = n or f(op(a[l], a[l + 1], ..., a[r])) = false.

        If f is monotone, this is the maximum r satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        assert 0 <= l <= self._n
        assert g(self.e)

        if l == self._n:
            return self._n

        l += self._size
        for i in range(self._log, 0, -1):
            self._push(l >> i)
        sm = self.e

        while True:
            while not l & 1:
                l >>= 1

            if not g(self.op(sm, self.tree[l])):
                while l < self._size:
                    l *= 2
                    if g(self.op(sm, self.tree[l])):
                        sm = self.op(sm, self.tree[l])
                        l += 1
                return l - self._size

            sm = self.op(sm, self.tree[l])
            l += 1

            if (l & -l) == l:
                break

        return self._n

    def min_left(self, r: int, g: Callable[[S], bool]) -> int:
        """
        Return an index l satisfying both:
            1. l = r or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. l = 0 or f(op(a[l - 1], a[l + 1], ..., a[r - 1])) = false.
        If f is monotone, this is the minimum l satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        assert 0 <= r <= self._n
        assert g(self.e)

        if not r:
            return 0

        r += self._size
        for i in range(self._log, 0, -1):
            self._push((r - 1) >> i)
        sm = self.e

        while True:
            r -= 1
            while r > 1 and r & 1:
                r >>= 1

            if not g(self.op(self.tree[r], sm)):
                while r < self._size:
                    r = 2 * r + 1
                    if g(self.op(self.tree[r], sm)):
                        sm = self.op(self.tree[r], sm)
                        r -= 1
                return r + 1 - self._size

            sm = self.op(self.tree[r], sm)

            if (r & -r) == r:
                break

        return 0


def range_affine_range_sum():
    """
    https://judge.yosupo.jp/problem/range_affine_range_sum
    https://atcoder.jp/contests/practice2/tasks/practice2_k
    """
    import sys

    readline = sys.stdin.buffer.readline

    N, Q = map(int, readline().split())
    A = [(i << 32) + 1 for i in map(int, readline().split())]

    MOD = 998244353
    e = 0
    id_ = 1 << 32

    def op(s, t):
        sv, sn = s >> 32, s % (1 << 32)
        tv, tn = t >> 32, t % (1 << 32)
        return (((sv + tv) % MOD) << 32) + sn + tn

    def mapping(f, a):
        fb, fc = f >> 32, f % (1 << 32)
        av, an = a >> 32, a % (1 << 32)
        return (((fb * av + fc * an) % MOD) << 32) + an

    def composition(f, g):
        fb, fc = f >> 32, f % (1 << 32)
        gb, gc = g >> 32, g % (1 << 32)
        return ((fb * gb % MOD) << 32) + ((gc * fb + fc) % MOD)

    tree = LazySegmentTree(A, e, op, id_, mapping, composition)
    res = []
    for _ in range(Q):
        com, l, r, *bc = map(int, readline().split())
        if com:
            res.append(tree.prod(l, r) >> 32)
        else:
            B, C = bc
            tree.apply_range(l, r, (B << 32) + C)
    print("\n".join(map(str, res)))


if __name__ == "__main__":
    range_affine_range_sum()

Solution (Python):
"""Buggy! the output of practice_k becomes about 1/4."""
from typing import Callable, List, TypeVar

S = TypeVar("S")
F = TypeVar("F")


class LazySegmentTree:
    """Lazy Segment Tree
    References:
        https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
    """

    __slots__ = [
        "e",
        "op",
        "id",
        "mapping",
        "composition",
        "_n",
        "_log",
        "_size",
        "tree",
        "lazy",
    ]

    def __init__(
        self,
        a: List[S],
        e: S,
        op: Callable[[S, S], S],
        id_: F,
        mapping: Callable[[F, S], S],
        composition: Callable[[F, F], F],
    ) -> None:
        self.e = e
        self.op = op
        self.id = id_
        self.mapping = mapping
        self.composition = composition

        self._n = len(a)
        self._log = (self._n - 1).bit_length()
        self._size = 1 << self._log

        self.tree = [e] * self._size + a + [e] * (self._size - self._n)
        for i in range(self._size - 1, 0, -1):
            self._update(i)

        self.lazy = [id_] * self._size

    def _update(self, k: int) -> None:
        """Update the value of a[k]."""
        self.tree[k] = self.op(self.tree[2 * k], self.tree[2 * k + 1])

    def _apply_all(self, k: int, f: F) -> None:
        self.tree[k] = self.mapping(f, self.tree[k])
        if k < self._size:
            self.lazy[k] = self.composition(f, self.lazy[k])

    def _push(self, k: int) -> None:
        self._apply_all(2 * k, self.lazy[k])
        self._apply_all(2 * k + 1, self.lazy[k])
        self.lazy[k] = self.id

    def set(self, k: int, x: S) -> None:
        """Assign x to a[k] in O(log n)."""
        

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        self.tree[k] = x
        for i in range(1, self._log + 1):
            self._update(k >> i)
        while k:
            k >>= 1
            self._update(k)

    def get(self, k: int) -> S:
        """Return a[k] in O(1)."""
        

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        return self.tree[k]

    def prod(self, l: int, r: int) -> S:
        """Return op(a[l], ..., a[r - 1]). Return e, if l == r.
        Complexity: O(log n)
        """
        

        if l == r:
            return self.e

        l += self._size
        r += self._size
        for i in range(self._log, 0, -1):
            if ((l >> i) << i) != l:
                self._push(l >> i)
            if ((r >> i) << i) != r:
                self._push(r >> i)

        sml, smr = self.e, self.e
        while l < r:
            if l & 1:
                sml = self.op(sml, self.tree[l])
                l += 1
            if r & 1:
                r -= 1
                smr = self.op(self.tree[r], smr)
            l >>= 1
            r >>= 1
        return self.op(sml, smr)

    def prod_all(self) -> S:
        """Return op(a[0], ..., a[n - 1]. Return e if n == 0.
        Complexity: O(1)
        """
        return self.tree[1]

    def apply(self, k: int, f: F) -> None:
        """Apply a[p] = op_st(a[p], x) in O(log n)."""
        

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        self.tree[k] = self.mapping(f, self.tree[k])
        for i in range(1, self._log + 1):
            self._update(k >> i)

    def apply_range(self, l: int, r: int, f: F) -> None:
        """Apply a[i] = op_st(a[i], x) for all i = l..r-1 in O(log n)."""
        

        if l == r:
            return

        l += self._size
        r += self._size
        for i in range(self._log, 0, -1):
            if ((l >> i) << i) != l:
                self._push(l >> i)
            if ((r >> i) << i) != r:
                self._push((r - 1) >> i)

        l_tmp, r_tmp = l, r
        while l < r:
            if l & 1:
                self._apply_all(l, f)
                l += 1
            if r & 1:
                r -= 1
                self._apply_all(r, f)
            l >>= 1
            r >>= 1
        l, r = l_tmp, r_tmp

        for i in range(1, self._log + 1):
            if ((l >> i) << i) != l:
                self._update(l >> i)
            if ((r >> i) << i) != r:
                self._update((r - 1) >> i)

    def max_right(self, l: int, g: Callable[[S], bool]) -> int:
        """
        Return an index r satisfying both:
            1. r = l or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. r = n or f(op(a[l], a[l + 1], ..., a[r])) = false.

        If f is monotone, this is the maximum r satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        
        

        if l == self._n:
            return self._n

        l += self._size
        for i in range(self._log, 0, -1):
            self._push(l >> i)
        sm = self.e

        while True:
            while not l & 1:
                l >>= 1

            if not g(self.op(sm, self.tree[l])):
                while l < self._size:
                    l *= 2
                    if g(self.op(sm, self.tree[l])):
                        sm = self.op(sm, self.tree[l])
                        l += 1
                return l - self._size

            sm = self.op(sm, self.tree[l])
            l += 1

            if (l & -l) == l:
                break

        return self._n

    def min_left(self, r: int, g: Callable[[S], bool]) -> int:
        """
        Return an index l satisfying both:
            1. l = r or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. l = 0 or f(op(a[l - 1], a[l + 1], ..., a[r - 1])) = false.
        If f is monotone, this is the minimum l satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        
        

        if not r:
            return 0

        r += self._size
        for i in range(self._log, 0, -1):
            self._push((r - 1) >> i)
        sm = self.e

        while True:
            r -= 1
            while r > 1 and r & 1:
                r >>= 1

            if not g(self.op(self.tree[r], sm)):
                while r < self._size:
                    r = 2 * r + 1
                    if g(self.op(self.tree[r], sm)):
                        sm = self.op(self.tree[r], sm)
                        r -= 1
                return r + 1 - self._size

            sm = self.op(self.tree[r], sm)

            if (r & -r) == r:
                break

        return 0


def practice2_k():
    
    import sys

    readline = sys.stdin.buffer.readline

    N, Q = map(int, readline().split())
    A = [(i << 32) + 1 for i in map(int, readline().split())]

    MOD = 998244353
    e = 0
    id_ = 1 << 32

    def op(s, t):
        sv, sn = s >> 32, s % (1 << 32)
        tv, tn = t >> 32, t % (1 << 32)
        return (((sv + tv) % MOD) << 32) + sn + tn

    def mapping(f, a):
        fb, fc = f >> 32, f % (1 << 32)
        av, an = a >> 32, a % (1 << 32)
        return (((fb * av + fc * an) % MOD) << 32) + an

    def composition(f, g):
        fb, fc = f >> 32, f % (1 << 32)
        gb, gc = g >> 32, g % (1 << 32)
        return ((fb * gb % MOD) << 32) + ((gc * fb + fc) % MOD)

    tree = LazySegmentTree(A, e, op, id_, mapping, composition)
    res = []
    for _ in range(Q):
        com, l, r, *bc = map(int, readline().split())
        if com:
            res.append(tree.prod(l, r) >> 32)
        else:
            B, C = bc
            tree.apply_range(l, r, (B << 32) + C)
    print("\n".join(map(str, res)))


if __name__ == "__main__":
    practice2_k()

Solution (Python):
"""Buggy! the output of practice_k becomes about 1/4."""
from typing import Callable, List, TypeVar

S = TypeVar("S")
F = TypeVar("F")


class LazySegmentTree:
    """Lazy Segment Tree
    References:
        https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
    """

    __slots__ = [
        "e",
        "op",
        "id",
        "mapping",
        "composition",
        "_n",
        "_log",
        "_size",
        "tree",
        "lazy",
    ]

    def __init__(
        self,
        a: List[S],
        e: S,
        op: Callable[[S, S], S],
        id_: F,
        mapping: Callable[[F, S], S],
        composition: Callable[[F, F], F],
    ) -> None:
        self.e = e
        self.op = op
        self.id = id_
        self.mapping = mapping
        self.composition = composition

        self._n = len(a)
        self._log = (self._n - 1).bit_length()
        self._size = 1 << self._log

        self.tree = [e] * self._size + a + [e] * (self._size - self._n)
        for i in range(self._size - 1, 0, -1):
            self._update(i)

        self.lazy = [id_] * self._size

    def _update(self, k: int) -> None:
        """Update the value of a[k]."""
        self.tree[k] = self.op(self.tree[2 * k], self.tree[2 * k + 1])

    def _apply_all(self, k: int, f: F) -> None:
        self.tree[k] = self.mapping(f, self.tree[k])
        if k < self._size:
            self.lazy[k] = self.composition(f, self.lazy[k])

    def _push(self, k: int) -> None:
        self._apply_all(2 * k, self.lazy[k])
        self._apply_all(2 * k + 1, self.lazy[k])
        self.lazy[k] = self.id

    def set(self, k: int, x: S) -> None:
        """Assign x to a[k] in O(log n)."""
        assert 0 <= k < self._n

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        self.tree[k] = x
        for i in range(1, self._log + 1):
            self._update(k >> i)
        while k:
            k >>= 1
            self._update(k)

    def get(self, k: int) -> S:
        """Return a[k] in O(1)."""
        assert 0 <= k < self._n

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        return self.tree[k]

    def prod(self, l: int, r: int) -> S:
        """Return op(a[l], ..., a[r - 1]). Return e, if l == r.
        Complexity: O(log n)
        """
        assert 0 <= l <= r <= self._n

        if l == r:
            return self.e

        l += self._size
        r += self._size
        for i in range(self._log, 0, -1):
            if ((l >> i) << i) != l:
                self._push(l >> i)
            if ((r >> i) << i) != r:
                self._push(r >> i)

        sml, smr = self.e, self.e
        while l < r:
            if l & 1:
                sml = self.op(sml, self.tree[l])
                l += 1
            if r & 1:
                r -= 1
                smr = self.op(self.tree[r], smr)
            l >>= 1
            r >>= 1
        return self.op(sml, smr)

    def prod_all(self) -> S:
        """Return op(a[0], ..., a[n - 1]. Return e if n == 0.
        Complexity: O(1)
        """
        return self.tree[1]

    def apply(self, k: int, f: F) -> None:
        """Apply a[p] = op_st(a[p], x) in O(log n)."""
        assert 0 <= k < self._n

        k += self._size
        for i in range(self._log, 0, -1):
            self._push(k >> i)
        self.tree[k] = self.mapping(f, self.tree[k])
        for i in range(1, self._log + 1):
            self._update(k >> i)

    def apply_range(self, l: int, r: int, f: F) -> None:
        """Apply a[i] = op_st(a[i], x) for all i = l..r-1 in O(log n)."""
        assert 0 <= l <= r <= self._n

        if l == r:
            return

        l += self._size
        r += self._size
        for i in range(self._log, 0, -1):
            if ((l >> i) << i) != l:
                self._push(l >> i)
            if ((r >> i) << i) != r:
                self._push((r - 1) >> i)

        l_tmp, r_tmp = l, r
        while l < r:
            if l & 1:
                self._apply_all(l, f)
                l += 1
            if r & 1:
                r -= 1
                self._apply_all(r, f)
            l >>= 1
            r >>= 1
        l, r = l_tmp, r_tmp

        for i in range(1, self._log + 1):
            if ((l >> i) << i) != l:
                self._update(l >> i)
            if ((r >> i) << i) != r:
                self._update((r - 1) >> i)

    def max_right(self, l: int, g: Callable[[S], bool]) -> int:
        """
        Return an index r satisfying both:
            1. r = l or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. r = n or f(op(a[l], a[l + 1], ..., a[r])) = false.

        If f is monotone, this is the maximum r satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        assert 0 <= l <= self._n
        assert g(self.e)

        if l == self._n:
            return self._n

        l += self._size
        for i in range(self._log, 0, -1):
            self._push(l >> i)
        sm = self.e

        while True:
            while not l & 1:
                l >>= 1

            if not g(self.op(sm, self.tree[l])):
                while l < self._size:
                    l *= 2
                    if g(self.op(sm, self.tree[l])):
                        sm = self.op(sm, self.tree[l])
                        l += 1
                return l - self._size

            sm = self.op(sm, self.tree[l])
            l += 1

            if (l & -l) == l:
                break

        return self._n

    def min_left(self, r: int, g: Callable[[S], bool]) -> int:
        """
        Return an index l satisfying both:
            1. l = r or f(op(a[l], a[l + 1], ..., a[r - 1])) = true
            2. l = 0 or f(op(a[l - 1], a[l + 1], ..., a[r - 1])) = false.
        If f is monotone, this is the minimum l satisfying:
            f(op(a[l], a[l + 1], ..., a[r - 1])) = true.

        Complexity: O(log n)
        """
        assert 0 <= r <= self._n
        assert g(self.e)

        if not r:
            return 0

        r += self._size
        for i in range(self._log, 0, -1):
            self._push((r - 1) >> i)
        sm = self.e

        while True:
            r -= 1
            while r > 1 and r & 1:
                r >>= 1

            if not g(self.op(self.tree[r], sm)):
                while r < self._size:
                    r = 2 * r + 1
                    if g(self.op(self.tree[r], sm)):
                        sm = self.op(self.tree[r], sm)
                        r -= 1
                return r + 1 - self._size

            sm = self.op(self.tree[r], sm)

            if (r & -r) == r:
                break

        return 0


def practice2_k():
    
    import sys

    readline = sys.stdin.buffer.readline

    N, Q = map(int, readline().split())
    A = [(i << 32) + 1 for i in map(int, readline().split())]

    MOD = 998244353
    e = 0
    id_ = 1 << 32

    def op(s, t):
        sv, sn = s >> 32, s % (1 << 32)
        tv, tn = t >> 32, t % (1 << 32)
        return (((sv + tv) % MOD) << 32) + sn + tn

    def mapping(f, a):
        fb, fc = f >> 32, f % (1 << 32)
        av, an = a >> 32, a % (1 << 32)
        return (((fb * av + fc * an) % MOD) << 32) + an

    def composition(f, g):
        fb, fc = f >> 32, f % (1 << 32)
        gb, gc = g >> 32, g % (1 << 32)
        return ((fb * gb % MOD) << 32) + ((gc * fb + fc) % MOD)

    tree = LazySegmentTree(A, e, op, id_, mapping, composition)
    res = []
    for _ in range(Q):
        com, l, r, *bc = map(int, readline().split())
        if com:
            res.append(tree.prod(l, r) >> 32)
        else:
            B, C = bc
            tree.apply_range(l, r, (B << 32) + C)
    print("\n".join(map(str, res)))


if __name__ == "__main__":
    practice2_k()