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;
using ll =long long;
typedef pair<int,int> P;
#define SORT(a) sort((a).begin(),(a).end())
#define REV(a) reverse((a).begin(),(a).end())
#define For(i, a, b)    for(int i = (a) ; i < (b) ; ++i)
#define rep(i, n)       For(i, 0, n)
#define debug(x)  cerr << #x << " = " << (x) << endl;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

void coY() {cout <<"Yes"<<endl;}
void coN(){cout <<"No"<<endl;}
 
//Write From this Line


const int dy[] = {0,0,1,-1};
const int dx[] = {1,-1,0,0};

//const ll mod = 1e9+7;
const ll mod = 998244353;
int main()
{
	ll n, k;
	cin >> n >> k;
	vector<int> l(k), r(k);
	rep(i,k) cin >> l[i] >> r[i];

	vector<ll> dp(200005,0);
	dp[1] = 1;
	dp[0] = 0;

	vector<ll> sum(200005,0);
	sum[0] = 0;
	sum[1] = 1;

	for(int i = 2; i <= n; i++){
		// dp[i] を更新する
		rep(j,k){
			ll right = i - l[j];
			if(right < 0) right = 0;
			ll left = i - r[j] - 1;
			if(left < 0) left = 0;
			ll tmp = sum[right] - sum[left];
			if(tmp < 0) tmp += mod;

			dp[i] += tmp;
			dp[i] %= mod;
		}
		sum[i] += sum[i-1] + dp[i];
		sum[i] %= mod;
	}
	cout << dp[n] << endl;
}