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>
#include <stdlib.h>
#include <fstream>
#include <iostream>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> llP;

ll mod(ll a, ll b){
  ll ret=a%b;
  if(ret<0) ret+=b;
  return ret;
}

ll modpow(ll a,ll b,ll c){
  ll res=1;
  while(b>0){
    if(b&1) res=mod(res*a,c);
    a=mod(a*a,c);
    b>>=1;
  }
  return res;
}

int GCD(int x,int y){
  if(x<y){
    swap(x,y);
  }
  while(x%y!=0){
    int temp=x;
    x=y;
    y=temp%y;
  }
  return y;
}

int digitsum(int x){
  int sum=0;
  while(x>0){
    sum+=(x%10);
    x/=10;
  }
  return sum;
}

int main() {
  ll MOD=998244353;
  ll n,k;
  cin>>n>>k;
  llP lr[k];
  for(int i=0;i<k;i++){
    ll l,r;
    cin>>l>>r;
    lr[i]=make_pair(l,r);
  }
  sort(lr,lr+k);
  ll sum[n],d[n];
  d[0]=1,sum[0]=1;
  for(int i=1;i<n;i++){
    d[i]=0;
    for(int j=0;j<k;j++){
      int a=lr[j].first,b=lr[j].second;
      if(a>i) break;
      if(b<i){
        d[i]+=(sum[i-a]-sum[i-b-1]);
        d[i]=mod(d[i],MOD);
      }
      if(b>=i){
        d[i]+=sum[i-a];
        d[i]=mod(d[i],MOD);
      }
    }
    sum[i]=sum[i-1]+d[i];
    sum[i]=mod(sum[i],MOD);
  }
  cout<<d[n-1]<<endl;
  return 0;
}