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 <stdio.h>

const int Mod = 998244353;

void add_BIT(int N, long long BIT[], int i, long long k)
{
	while (i <= N) {
		BIT[i] = (BIT[i] + k) % Mod;
		i += (i & -i);
	}
}

long long sum_BIT(long long BIT[], int r)
{
	long long sum = 0;
	while (r > 0) {
		sum = (sum + BIT[r]) % Mod;
		r -= (r & -r);
	}
	return sum;
}

int main()
{
	int i, N, K, L[11], R[11];
	scanf("%d %d", &N, &K);
	for (i = 1; i <= K; i++) scanf("%d %d", &(L[i]), &(R[i]));
	
	int j;
	long long BIT[200001] = {}, dp[200001] = {0, 1};
	add_BIT(N, BIT, 1, 1);
	for (i = 2; i <= N; i++) {
		for (j = 1; j <= K; j++) {
			if (L[j] < i) {
				dp[i] += sum_BIT(BIT, i - L[j]);
				if (R[j] < i) dp[i] += Mod - sum_BIT(BIT, i - R[j] - 1);
			}
		}
		dp[i] %= Mod;
		add_BIT(N, BIT, i, dp[i]);
	}
	printf("%lld\n", dp[N]);
	fflush(stdout);
	return 0;
}