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)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define pb(x)           push_back(x)
#define mkp(x,y)        make_pair(x,y)
#define L               length()
#define int             long long
#define bs              binary_search
#define mod             1e9+7
#define IOS             ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define all(v)          v.begin(),v.end()
#define allr(v)         v.rbegin(),v.rend()
#define mem0(a)         memset(a,0,sizeof(a))
#define mem1(a)         memset(a,-1,sizeof(a))
#define F               first
#define S               second
#define pii             pair<int,int>
#define vi              vector<int>
#define vs              size()
#define endl            '\n'
#define atoi            stoi
#define elasped_time    1.0 * clock() / CLOCKS_PER_SEC
#define si              set <int>
#define vpii            vector < pair <int,int> >
#define memf(a)         memset(a,false,sizeof(a))
#define memt(a)         memset(a,true,sizeof(a))
#define xxx             998244353
#define pi              3.141592653589
#define ninf            INT_MIN
#define inf             INT_MAX
#define sz(v)           ((int)(v).size())
#define rep(i,a,b)      for(int i=a;i<=b;i++)
int max(int a,int b){if(a>b){return a;}else{return b;}}
int min(int a,int b){if(a<b){return a;}else{return b;}}
int power(int b,int e)
{
    if(e==0)
    return 1;
    if(e%2)
    return ((b*power((b)*(b),(e-1)/2)));
    else
    return power((b)*(b),e/2);
}
int modpower(int b,int e,int q)
{
    int MOD=q;
    if(e==0)
    return 1;
    if(e%2)
    return ((b%MOD)*modpower((b%MOD)*(b%MOD),(e-1)/2,q))%MOD;
    else
    return modpower((b%MOD)*(b%MOD),e/2,q)%MOD;
}
void dpv(vi v)
{
    for(int i=0;i<v.vs;i++)
    {
        cout<<v[i]<<" ";
    }
    cout<<endl;
}
void dpv(vpii v)
{
    for(int i=0;i<v.vs;i++)
    {
        cout<<v[i].F<<" "<<v[i].S<<endl;
    }
}
void dpv(set <int> v)
{
    for(auto i:v)
    {
        cout<<i<<" ";
    }
    cout<<endl;
}
//////////////////////**TREE MOVES STARTS**//////////////////////////////
///// **TREE MOVES ENDS**////////////////////////////

////*SOLUTION TO THE QUESTION STARTS HERE*///////
vi pref(200001,0);
vi dp(200001,0);
int sum(int l,int r)
{
    l=max(l,0);
    return (pref[r]-(l==0?0:pref[l-1]));
}
void oblivious()
{
    int m1=998244353;
    int n,k;
    cin>>n>>k;
    int l[k],r[k];
    pref[1]=1;
    dp[1]=1;
    for(int i=0;i<k;i++)
    {
        cin>>l[i]>>r[i];
    }
    for(int i=2;i<=n;i++)
    {
        for(int j=0;j<k;j++)
        {
            if(i-l[j]>=0)
            {
                int cur=sum(i-r[j],i-l[j])%m1;
                dp[i]+=cur%m1;
                dp[i]%=m1;
            }
        }
        pref[i]=(pref[i-1]%m1+dp[i]%m1)%m1;
    }
    dp[n]%=m1;
    cout<<(dp[n]+m1)%m1<<endl;
}
signed main()
{
    IOS;
    /*#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif*/
    int t=1;
    //cin>>t;
    //int z=t;
    while(t--)
    {
        //cout<<"Case #"<<(z-t)<<": ";
        oblivious();
    }
    return 0;
}
//Editorial
/*

    */