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: Wrong Answer)

#include <stdio.h>
int main()
{
  int n,k;
  scanf("%d%d",&n,&k);
  int i = 0;
  int l[k],r[k];
  unsigned long s[200001] = {0};
  int idx = 0;
  int tmp;
  while (i < k)
  {
    scanf("%d%d",&l[i],&r[i]);
    tmp = l[i];
    while (tmp <= r[i])
    {
      s[idx] = tmp;
      tmp++;
      idx++;
    }
    i++;
  }
  unsigned long sum = 1;
  i = 0;
  int j;
  int ans;
  int cnt;
  /*
  ** mod(n,sの要素)が0になるまで要素をずらしながら繰り返すことで
  ** sの和集合から任意に選択してnまで加算する組み合わせを出せる気がする
  ** 加算してnになるパターン一つを検出すると、その順序すべてをカウントできる、はず
  ** 余りだけでなく、商がないと順序の組み合わせを出せない
  ** 商と余りをセットで格納する必要があり、また配列作って～は非常にめんどい
  */
  while (s[i] != 0)
  {
    j = i;
    ans = -1;
    while ((ans = n % s[j]) && s[j] != 0)
    {
      j++;
    }
    if (ans == 0)
    {
      
    }
    i++;
  }
  
}