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; 
typedef long long ll;
const ll mxn = 1e5;
const ll mod = 998244353;
 
ll pow(ll a,ll b){
    ll r =1;
    while(b){
        if(b%2!=0) r=(r*a)%mod;
        a=(a*a)%mod;
        b/=2;
    }
    return r;
}
 
const int siz = 5e5+20;
ll fact[siz], inv[siz];
void pre(){
	fact[0] = 1;
	inv[0] = pow(fact[0], mod-2);
	for(int i=1;i<siz;i++){
		fact[i] = (fact[i-1]*i)%mod;
	}
}
 
int spf[10];
void sieve(){
    spf[1]=1;
    for(int i=2;i<mxn;i++) spf[i]=i;
    for(int i=4;i<mxn;i+=2) spf[i]=2;
    for(int i=3;i*i<mxn;i++){
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll n,k;
    cin>>n>>k;

    vector<pair<ll,ll>> v;

    for(int i=0;i<k;i++){
        ll a,b;
        cin>>a>>b;
        v.push_back(make_pair(a,b));
    }

    sort(v.begin(),v.end());

    ll dp[n];
    memset(dp,0,sizeof(dp));
    dp[0]=1;

    for(int i=1;i<n;i++){
        for(int j=0;j<v.size();j++){
            for(int k=v[j].first;k<=v[j].second;k++){
                if(k>i) break;
                dp[i] = (dp[i]+dp[i-k])%mod;
            }
            if(v[j].second>i) break;
        }
    }

    cout<<dp[n-1];
}