Problem:
Score : 400 points

Problem Statement

There are N cells arranged in a row, numbered 1, 2, \ldots, N from left to right.

Tak lives in these cells and is currently on Cell 1. He is trying to reach Cell N by using the procedure described below.

You are given an integer K that is less than or equal to 10, and K non-intersecting segments [L_1, R_1], [L_2, R_2], \ldots, [L_K, R_K].
Let S be the union of these K segments.
Here, the segment [l, r] denotes the set consisting of all integers i that satisfy l \leq i \leq r.

When you are on Cell i, pick an integer d from S and move to Cell i + d. You cannot move out of the cells.

To help Tak, find the number of ways to go to Cell N, modulo 998244353.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq K \leq \min(N, 10)

1 \leq L_i \leq R_i \leq N

[L_i, R_i] and [L_j, R_j] do not intersect (i \neq j)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
L_1 R_1
L_2 R_2
:
L_K R_K

Output

Print the number of ways for Tak to go from Cell 1 to Cell N, modulo 998244353.

Sample Input 1

5 2
1 1
3 4

Sample Output 1

4

The set S is the union of the segment [1, 1] and the segment [3, 4], therefore S = \{ 1, 3, 4 \} holds.

There are 4 possible ways to get to Cell 5:

1 \to 2 \to 3 \to 4 \to 5,

1 \to 2 \to 5,

1 \to 4 \to 5 and

1 \to 5.

Sample Input 2

5 2
3 3
5 5

Sample Output 2

0

Because S = \{ 3, 5 \} holds, you cannot reach to Cell 5.
Print 0.

Sample Input 3

5 1
1 2

Sample Output 3

5

Sample Input 4

60 3
5 8
1 3
10 15

Sample Output 4

221823067

Note that you have to print the answer modulo 998244353.

Solution in Python (3.8.2) (judge verdict: Time Limit Exceeded)

import sys
sys.setrecursionlimit(300000)

N, K = map(int, input().split())
L = 998244353

S_tmp = {}
for _ in range(K):
    l, r = map(int, input().split())
    S_tmp.update({ l : r })
#S = []
S_ = []
for l, r in sorted(S_tmp.items(), key=lambda x:x[0]):
    #S.extend( list(range(l,r+1)) )
    S_.append((l,r))


import numpy as np

A = np.array([0 for _ in range(N+1)])

#from functools import lru_cache
#@lru_cache(maxsize=None)

A[1] = 1
for n in range(2,N+1):
    A[n] = sum([ A[max(n-r,1):n-l+1].sum() % L for (l,r) in S_ if l < n ]) % L

print(A[n] % L)

def compute(n):
    if n == 1:
        return 1
    a = 0
    for s in S:
        if s >= n:
            break
        if A[n-s] is None:
            A[n-s] = compute(n-s)
        a += A[n-s]
    return a % L

#print(compute(N) % L)