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>
typedef long long int lli;

#define em emplace_back
#define rep(i,j,n) for(lli i=j;i<n;i++)
#define repf(i,a,j,n) for(lli i=j,a=0;i<n;i++,a++)
#define repg(i,j,n) for(int i=j;i>n;--i)
using namespace std;
lli m = 998244353;

void solve(){

    lli n,k;
    cin>>n>>k;

    vector<lli> v;

    vector<lli> arrfront(n+1,0);

    unordered_map<lli,lli> s;
    
    rep(i,0,k){
        lli a,b;
        cin>>a>>b;
        if(a==b){
            v.em(a);
            continue;
        }
        rep(j,a,b+1){
            v.em(j);
        }
    }

    vector<lli> arrback(n+1,0);
    sort(v.begin(), v.end());
    lli ans = 0;
    rep(i,1,n){

        if(arrfront[i]==0){
            if(i!=1){
                continue;
            }else{
                arrfront[i] = 1;
            }
        }

        rep(j,0,v.size()){
            if(i+v[j]>n){
                break;
            }
            arrfront[i+v[j]] = arrfront[i+v[j]]%m;
            arrfront[i+v[j]] = (arrfront[i+v[j]]%m + arrfront[i]%m)%m;
            arrfront[i+v[j]] = arrfront[i+v[j]]%m; 
            
        }
    }

    cout<<(arrfront[n]%m);

}
 
 
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
 
    return 0;
}