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: Runtime Error)

#include<stdio.h>
int main()
{
	long long int n, k;
	scanf("%lld %lld", &n, &k);
	long long int i, j;
	long long int l[11], r[11];
	for (i = 0; i < k; i++)
		scanf("%lld %lld", &l[i], &r[i]);
	long long int p = 998244353;
	long long int cnt[200005], d[200005];
	for (i = 0; i <= n; i++)
		cnt[i] = d[i] = 0;
	cnt[0] = 1;
	for (i = 0; i < n; i++)
	{
		if (i > 0)
		{
			d[i] = (d[i] + d[i - 1]) % p;
			cnt[i] = d[i];
		}
		for (j = 0; j < k; j++)
		{
			d[l[j] + i] = (d[l[j] + i] + cnt[i]) % p;
			d[r[j] + 1 + i] = (d[r[j] + 1 + i] - cnt[i] + p) % p;
		}
	}
	printf("%lld\n", cnt[n - 1]);
	return 0;
}