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: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>
#define MOD 998244353
int count(int n);
int a[200010] = {};
int b[200010] = {};
int main()
{
    int n, k;
    int ans = 0;
    int l, r;
    scanf("%d %d", &n, &k);
    for (int i = 0; i < k; i++)
    {
        scanf("%d %d", &l, &r);
        for (int i = l; i <= r; i++)
        {
            a[i] = 1;
        }
    }
    printf("%d", count(n - 1));
    return 0;
}

int count(int t)
{
    if (b[t] == -1)
    {
        return 0;
    }
    if (b[t] == 0)
    {
        b[t] = a[t];
        for (int i = 1; i < t; i++)
        {
            if (a[i] == 1)
            {
                b[t] += count(t - i);
                b[t] %= MOD;
            }
        }
    }
    if (b[t] == 0)
    {
        b[t] = -1;
        return 0;
    }
    return b[t];
}