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)

/*
 * FileName:     D
 * CreatedDate:  2020-09-19 21:57:10 +0900
 * LastModified: 2020-09-19 22:07:03 +0900
 */

#include <stdio.h>
#include <string.h>
int main(void){
    int N, K;scanf("%d%d", &N, &K);
    int box[K][2];
    for(int i=0; i<K; i++){
        scanf("%d%d", &box[i][0], &box[i][1]);
    }
    int dp[N+1];
    int unko = 998244353;
    memset(dp, 0, sizeof(dp));
    dp[1] = 1;
    for(int i=1; i<N+1; i++){
        if(dp[i] != 0){
            for(int j=0; j<K; j++){
                for(int k=box[j][0]; k<box[j][1]+1; k++){
                    if(i+k>N){
                        break;
                    }
                    dp[i+k] += dp[i];
                    dp[i+k] %= unko;
                }
            }
        }
    }
    printf("%d\n", dp[N] % unko);
    return 0;
}