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>
#define maxN 200010

const long long int mod = 998244353, def = -1;
long long int n, k, l, r, numS = 0, S[maxN] = { 0 }, ans[maxN] = { 0 }, DP[maxN] = { def };

long long int dp(long long int a) {
	long long int p = a;
	if (p < 1) return 0;
	if (p == 1) return 1;

	long long int rtn = 0;
	for (int i = 0; i < numS; i++) {
		long long int N = p - S[i];
		long long int plus;
		if (DP[N] != def)plus = DP[N];
		else {
			plus = dp(N);
			DP[N] = plus;
		}
		rtn += plus;
      	rtn = rtn % mod;
		//printf("\ndp:%d %d", N, dp(N));
        //for(int j=1;j<=n;j++)printf("%d",DP[j]);
	}
	return rtn;
}

int main() {
	scanf("%lld%lld", &n, &k);
	for (int i = 0; i < k; i++) {
		scanf("%lld%lld", &l, &r);
		for (int j = l; j <= r; j++) {
			S[numS] = j;
			numS++;
		}
	}
	for (int i = 1; i <= n; i++)DP[i] = def;
	long long int ans = dp(n);
	printf("%lld", ans);
	return 0;
}