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 <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define FOR(i,l,r) for(i=l;i<r;i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(x) x.begin(),x.end()
#define P pair<ll,ll>
#define F first
#define S second
vector<ll>S;ll mod=998244353;
ll solve(ll SUM,vector<ll>&X){
  if(X.at(SUM)==-1){
    ll ans=0;
    for(ll i=0;i<S.size()&&S.at(i)<=SUM;i++){
      ans+=solve(SUM-S.at(i),X);ans%=mod;if(X.at(SUM)!=-1)break;
    }
    if(X.at(SUM)==-1)X.at(SUM)=ans%mod;
  }
  return X.at(SUM);
}
signed main(){
  ll N,K,i,j,L,R;cin>>N>>K;vector<ll>X(N,-1);
  REP(i,K){cin>>L>>R;FOR(j,L,R+1)S.push_back(j);}sort(ALL(S));
  REP(i,S.at(0))X.at(i)=0;X.at(i)=1;X.at(0)=1;
  cout<<solve(N-1,X)<<endl;
  return 0;
}