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: Runtime Error)

#include<bits/stdc++.h>
#define mp make_pair
#define fi first
#define se second
#define VAL 100000
#define ll  long  long
#define INF 0x3f3f3f3f
#define pll pair<long long , long long>
const int MAX=1e6+9;
const ll mod=998244353;
const int TOT_PRIMES=19;
const int MAX_A=70;
using namespace std;
vector<int>adj[MAX];
int prime[MAX];
int main() 
{
	
	int n , k;
	cin>>n>>k;
	ll dp[n+10000];
	memset(dp,0, sizeof(dp));
	dp[0]=0;
	dp[1]=1;
	vector<pair<int,  int > > v;
	for(int i=0;i<k;i++)
	{
		int x, y;
		cin>>x>>y;
		v.push_back({x,y});
	}
	for(int i =1;i<=n+5;i++){
		if(i>2)
			dp[i]+=dp[i-1];

		ll now = dp[i]%mod;
		for(int j =0;j<v.size();j++){
			dp[i+v[j].first]=(dp[i+v[j].first]+now);
			dp[i+v[j].second+1]=(dp[i+v[j].second+1]-now);
		}
        //cout<<dp[i]<<endl;
	}
	cout<<dp[n]%mod;
	
	
}