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: Time Limit Exceeded)

#include "bits/stdc++.h"
using namespace std;
#pragma GCC optimize "trapv"
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define pb push_back
#define fi first
#define se second
#define all(container) container.begin(),container.end()
#define fast ios::sync_with_stdio(0); cin.tie(0)
#ifndef ONLINE_JUDGE
#include "debug.cpp"
#endif
#define int long long
const int mod = 998244353;
int power(int a,int b) { if(b==0) return 1; int res = power(a,b/2); if(b&1) return res * res * a; else return res * res; }

const int sz = 200000+1;
int dp[sz];
vector<int> v;

int go(int pos,int n){
  if(dp[pos]!=-1){
    return dp[pos];  
  }else if(pos==n){
    return 1;
  }else{
    int ans = 0;
    for(int i=0;i<(int)v.size();i++){
      if(pos+v[i]<=n) ans = (ans%mod + go(pos+v[i],n)%mod)%mod;
    }
    dp[pos] = ans;
    return dp[pos];
  }
}

int32_t main(){
    
  fast;
  memset(dp,-1,sizeof(dp));

  int n,k; cin >> n >> k;
  
  int diff[200000+1] = {0};

  for(int i=1;i<=k;i++){
    int l,r; cin >> l >> r;
    diff[l]++;
    diff[r+1]--;
  }

  for(int i=1;i<n+1;i++){
    diff[i] = diff[i] + diff[i-1];
    if(diff[i]!=0) v.pb(i);
  }

  
  cout << go(1,n) << '\n'; 
  return 0;
}