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: Accepted)

#include<iostream>
#include<vector>
using LL = long long;
using namespace std;
int mod = 998244353;
int main() {
	LL N, K;
	cin >> N >> K;
	vector<LL> Li(K), Ri(K);
	vector<bool> S_has(N + 1, false);
	for (int i = 0; i < K; i++) {
		cin >> Li[i] >> Ri[i];
		for (int j = Li[i]; j <= Ri[i]; j++) {
			S_has[j] = true;
		}
	}
	vector<LL> Dp(N + 1, 0);
	Dp[1] = 1;

	for (int i = 2; i <= N; i++) {
		Dp[i] += Dp[i - 1];
		for (int k = 0; k < K; k++) {
			if (i - Li[k] > 0) {
				Dp[i] += Dp[i - Li[k]];
			}
			if (i - Ri[k] - 1 > 0) {
				Dp[i] -= Dp[i - Ri[k] - 1];
			}
		}
		while (Dp[i] < 0) {
			Dp[i] += mod;
		}
		Dp[i] = Dp[i] % mod;

	}
	LL ans = Dp[N] - Dp[N - 1];
	while (ans < 0) {
		ans += mod;
	}
	cout << ans << endl;

	return 0;
}