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>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <inttypes.h>

int compare_int(const void *a, const void *b){//qsort 整数比較<
	return *(int*)a - *(int*)b;
}
int compare_int2(const void *a, const void *b){//qsort 整数比較>
	return *(int*)b - *(int*)a;
}


//long long int combination(int n, int r)
//{
//	if (r == 0 || r == n)
//		return (1);
//	else if (r == 1)
//		return (n);
//	return (combination(n - 1, r - 1) + combination(n - 1, r));
//}

int main(void){
	int n,k;
	long long int ans[200010]={0,1,0};
	long long int s[200010]={0};
	int l[15]={0},r[15]={0};
	int i,j;
	int count=0;


	scanf("%d %d",&n,&k);
	for(i=0;i<k;i++){
		scanf("%d %d",&l[i],&r[i]);
	}
	qsort(l,k,sizeof(int),compare_int);
	qsort(r,k,sizeof(int),compare_int);
	for(i=0;i<k;i++){
		for(j=l[i];j<=r[i];j++){
			s[count]=j;
			count++;
		}
	}
	for(i=2;i<=n;i++){
		for(j=0;j<count;j++){
			if(i>s[j]){
				ans[i]+=ans[i-s[j]];
			}
			ans[i]=ans[i]%998244353;
		}
	}
	printf("%lld",ans[n]);
	return 0;
}