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<bits/stdc++.h>
using namespace std;


using ll =long long;
const ll MOD = 998244353;

int main(){
    ll n,i,k,j;
    cin>>n>>k;
    ll dp[n+2] ={0};
    dp[1] = 0;
    ll arr[k][2];
    for(i=0;i<k;i++)cin>>arr[i][0]>>arr[i][1];
    for(i=0;i<k;i++){
        dp[min(n+1,1+arr[i][0])]++;
        dp[min(n+1,1+arr[i][1]+1)]--;
    }

    //for(i=1;i<=n+1;i++)cout<<dp[i]<<" ";
    //cout<<"\n";
    for(i=2;i<=n;i++){
        //for(int q=1;q<=n+1;q++)cout<<dp[q]<<" ";
        //cout<<"\n";
        dp[i]+=dp[i-1];
        while(dp[i]<0)dp[i]+=MOD;
        dp[i]%=MOD;
        for(j=0;j<k;j++){
            dp[min(n+1,i+arr[j][0])]+=dp[i];
            dp[min(n+1,i+arr[j][1]+1)]-=dp[i];
        }
        //cout<<dp[i]<<"\n";
    }
    //for(i=1;i<=n+1;i++)cout<<dp[i]<<" ";
    //dp[n]+=dp[n-1];
    cout<<(dp[n]%MOD);
}