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: Runtime Error)

#include<stdio.h>

void quick_sort(int *array, int left, int right)   //left, right: ソート対象の要素の左端と右端の添字
{
    int i, j, n;
    int left_key=left, right_key=right;
    int pivot=array[(left+right)/2];    //ピボットの設定

    while(1)
    {
        while(array[left_key]<pivot)
        {
            left_key++;
        }
        while(array[right_key]>pivot)
        {
            right_key--;
        }
        if(left_key>=right_key)
        {
            break;
        }
        n=array[left_key];
        array[left_key]=array[right_key];
        array[right_key]=n;

        left_key++;
        right_key--;
    }    

    if((left_key-left)>1)   //左側が2個以上の時
    {
        quick_sort(array, left, left_key-1);
    }
    if ((right-right_key)>1)   //右側が2個以上の時
    {
        quick_sort(array, right_key+1, right);
    }
}

int main()
{
    int scnt=0/*データ数-1*/,x,y,i,j,k,N,K;
    scanf("%d %d",&N,&K);
    int S[N],dp[N]/*dp:各マスへの移動方法の総数 */;
    for(i=0;i<K;i++)
    {
        scanf("%d %d",&x,&y);
        for(j=x;j<=y;j++)
        {
            S[scnt]=j;
            scnt++;
        }
    }
    quick_sort(S,0,scnt-1);

    for(i=0;i<N;i++)
    {
        dp[i]=0;
    }
    for(i=0;i<scnt;i++)
    {
        dp[S[i]]++;
    }

    for(i=0;i<N;i++)
    {
        for(j=0;j<scnt;j++)
        {
            dp[i+S[j]]=dp[i+S[j]]+dp[i];
        }
    }

    printf("%d",dp[N-1]);
    return 0;
}