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)

#bitでタコ殴りにする
class Bit:
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)
        self.depth = n.bit_length()
 
    def sum(self, i):
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
 
    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i
 
    def lower_bound(self, x):
        """ 累積和がx以上になる最小のindexと、その直前までの累積和 """
        sum_ = 0
        pos = 0
        for i in range(self.depth, -1, -1):
            k = pos + (1 << i)
            if k <= self.size and sum_ + self.tree[k] < x:
                sum_ += self.tree[k]
                pos += 1 << i
        return pos + 1, sum_

m=998244353    
N,K=map(int,input().split())
bit = Bit(N)
bit.add(1,1)#初期値　はじめは１
period = []
for _ in range(K):#動ける区間の記憶
    l,r=map(int,input().split())
    period.append((l,r))
for i in range(2,N+1):#2からNまで見ていく。もらうDP
    for l,r in period:
        rx = max(0,i-l)
        lx = max(0,i-r-1)
        if rx==0:
            tmp_r = 0
        else:
            tmp_r = bit.sum(rx)
        if lx == 0:
            tmp_l = 0
        else:
            tmp_l = bit.sum(lx)
        bit.add(i,(tmp_r-tmp_l)%m)
ans = (bit.sum(N)-bit.sum(N-1))%m
print(ans)