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>

#define MOD_BY 998244353

int add(int a, int b) {
	return a + b - MOD_BY * (a + b >= MOD_BY);
}

int sub(int a, int b) {
	return b == 0 ? a : add(a, MOD_BY - b);
}

#define KI_MAX 212345

int ki[KI_MAX];

void ki_add(int idx, int value) {
	idx++;
	while (idx < KI_MAX) {
		ki[idx] = add(ki[idx], value);
		idx += idx & -idx;
	}
}

int ki_sum(int idx) {
	int ret = 0;
	if (idx < 0) return 0;
	idx++;
	if (idx >= KI_MAX) idx = KI_MAX - 1;
	while (idx > 0) {
		ret = add(ret, ki[idx]);
		idx -= idx & -idx;
	}
	return ret;
}

/* [s, e] */
int ki_sum2(int s, int e) {
	return sub(ki_sum(e), ki_sum(s - 1));
}

int N, K;
int L[16], R[16];

int main(void) {
	int i;
	if (scanf("%d%d", &N, &K) != 2) return 1;
	for (i = 0; i < K; i++) {
		if (scanf("%d%d", &L[i], &R[i]) != 2) return 1;
	}
	ki_add(N, 1);
	for (i = N - 1; i >= 1; i--) {
		int mofu = 0;
		int j;
		for (j = 0; j < K; j++) {
			mofu = add(mofu, ki_sum2(i + L[j], i + R[j]));
		}
		ki_add(i, mofu);
	}
	printf("%d\n", ki_sum2(1, 1));
	return 0;
}