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 PyPy3 (7.3.0) (judge verdict: Accepted)

import sys
input = sys.stdin.readline

n, k = map(int,input().split())
C = [list(map(int,input().split())) for i in range(k)]

# for i in range(k):
#     C[i][0] -= 1
#     C[i][1] -= 1

# ACLに伴い0-indexedに直した
class Bit:  # Fenwick Tree と同じ
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)
  
    def sum(self, l, r):  # [l,r)の合計を求める
        # 内部的には[l+1,r+1)の合計つまり(rまでの合計-lまでの合計)
        s = 0
        while r > 0:
            s += self.tree[r]
            r -= r & -r
        while l > 0:
            s -= self.tree[l]
            l -= l & -l
        return s
  
    def add(self, i, x):  # i番目にxを足す
        i += 1
        while i <= self.size:
            self.tree[i] += x
            i += i & -i

# Bitの内部状態をindex順に出力
def print_bit(B):
    n = B.size
    ANS = []
    for i in range(n):
        ANS.append(B.sum(i, i+1))
    print(*ANS)

B = Bit(n)
mod = 998244353
B.add(0, 1)
for i in range(1, n):
    for j in range(k):
        B.add(i, (B.sum(max(0,i-C[j][1]), max(0,i-C[j][0]+1)))%mod)

print(B.sum(n-1,n)%mod)