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>
#include <stdlib.h>

int fun (int a, int n) {
    if (a >= 0) return a;
    return n;
}

int main (void) {
    int n, k;
    scanf("%d %d", &n, &k);

    int lr[n][2], i, j;
    long int a, sum[n + 1];
    sum[0] = 1;
    sum[n] = 0;
    for (i = 0; i < k; i++) {
        scanf("%d %d", &lr[i][0], &lr[i][1]);
        lr[i][1]++;
    }

    for (i = 1; i < n; i++) {
        a = 0;
        for (j = 0; j < k; j++) {
            if (i - lr[j][0] >= 0)
                a += sum[i - lr[j][0]] - sum[fun(i - lr[j][1], n)];
        }
        sum[i] = (sum[i - 1] + a) % 998244353;
    }

    if (sum[n - 1] < sum[n - 2]) sum[n - 1] += 998244353;
    printf("%ld\n", sum[n - 1] - sum[n - 2]);
    return 0;
}