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)

//                        Bismillahir Rahmanir Rahim
//ALGO:
#include<bits/stdc++.h>
#define pb push_back
#define F first
#define S second
#define all(v) v.begin(), v.end()
#define FILL(a, x) memset(a, x, sizeof(a))
#define ll long long
#define Fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define INF INT_MAX
#define MX 1000010
#define mod  998244353
#define dout if(debug) cout
#define FR(i, n) for(int i=0; i<n; i++)
#define FOR(i, n) for(int i=1; i<=n; i++)
const double Pi=acos(-1);
using namespace std;
int debug=01;
void solve(){
    ll n, k;
    cin>>n>>k;
    ll x[k], y[k], cnt[n+1]={0};
    for(ll i=0; i<k; i++)cin>>x[i]>>y[i];
    cnt[1]=1;
    ll sum=0;
    for(ll i=2; i<=n; i++){
        for(ll j=0; j<k; j++){
            if(i-x[j]>0)sum=(sum+cnt[i-x[j]])%mod;
            if(i-y[j]-1>0)sum=(sum-cnt[i-y[j]-1]+mod)%mod;
        }
        cnt[i]=sum;
    }
    cout<<cnt[n];
}
int main() {
    Fast
    int tst=1;
    //cin>>tst;
    for(int T=1; T<=tst; T++){
        //cout<<"Case "<<T<<": ";
        solve();
    }
}