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 C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>

#define LIMIT 998244353

int main(){
   int  i,j,k,n,s[200001],l,r,d[200001];
   long a[200001],t;

   scanf("%d %d",&n,&k);
   for (i=1;i<=n;i++) s[i]=0;
   for (i=0;i<k;i++) {
      scanf("%d %d",&l,&r);
      for (j=l;j<=r;j++) s[j]=1;
   }
   for (j=n;j>0 && !s[j];j--) ;
   for (i=n;i>0;i--) {
      if (i<=j)
         for (j=i-1;j>0 && !s[j];j--) ;
      d[i]=i-j;
   }
   a[1]=1;
   for (i=2;i<=n;i++) {
      t=0;
      for (j=1;j<i;j+=d[i-j])
         if (s[i-j]) t=(t+a[j])%LIMIT;
      a[i]=t;
   }
   printf("%ld\n",a[n]);
   return 0;
}