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>
using namespace std;
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> p2;
#define sz(a) ll(a.size())
ll t[800005],hell=998244353;
ll add(ll x,ll y){
	x+=y;
	if(x>=hell)
		x-=hell;
	return x;
}
void change(ll v,ll tl,ll tr,ll pos,ll val){
	if(tl==tr){
		t[v]=val;
		return;
	}
	ll tm=(tl+tr)/2;
	if(pos<=tm)
		change(2*v,tl,tm,pos,val);
	else
		change(2*v+1,tm+1,tr,pos,val);
	t[v]=add(t[2*v],t[2*v+1]);
}
ll get(ll v,ll tl,ll tr,ll l,ll r){
	if(l>r)
		return 0;
	if(l==tl&&r==tr)
		return t[v];
	ll tm=(tl+tr)/2;
	return add(get(2*v,tl,tm,l,min(r,tm)),get(2*v+1,tm+1,tr,max(l,tm+1),r));
}
void solve(){
	ll n,k;
	cin>>n>>k;
	vector<p2> v(k);
	for(ll i=0;i<k;i++)
		cin>>v[i].first>>v[i].second;
	change(1,1,n,n,1);
	for(ll i=n-1;i>=1;i--){
		ll res=0;
		for(ll j=0;j<k;j++){
			ll l=i+v[j].first,r=min(i+v[j].second,n);
			if(l>r)
				continue;
			res=add(res,get(1,1,n,l,r));
		}
		change(1,1,n,i,res);
	}
	cout<<get(1,1,n,1,1);
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r" , stdin);
    freopen("output.txt", "w", stdout);
    #endif
    clock_t z=clock();
    ll qc=1;
    //cin>>qc;
    for(ll i=1;i<=qc;i++){
        solve();
    }
    debug("Total Time:%.4Lf\n",(ld)(clock()-z)/CLOCKS_PER_SEC);
}