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 ll = long long;
using namespace std;
const int INFint = 1e9+1;
const ll INFll = (ll)1e18+1;
ll MOD=998244353;


int main(){
  ll N,K;
  cin>>N>>K;
  vector<ll> L(K),R(K);
  for(int i(0);i<K;i++){
    cin>>L[i]>>R[i];
  }
  ll dp[200001] = {0}, r[200001] = {0};
  dp[1] = 1;
  ll tmp(0);
  for(int i(1);i<=N;i++){
    tmp += r[i]; tmp %= MOD;
    if(tmp < 0) tmp += MOD;
    if(i != 1) dp[i] = tmp;
    for(int k(0);k<K;k++){
      if(i+L[k] <= N){
        r[i+L[k]] += dp[i];
        r[i+L[k]] %= MOD;
      }
      if(i+R[k]+1 <= N){
        r[i+R[k]+1] -= dp[i];
        //if(r[i+R[k]+1] < 0) dp[]
      }
    } 
  }
  cout << dp[N] << endl;
  return 0;
}