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;
#define rep(i,n)for(int i=0;i<(int)(n);i++)
#define PI 3.141592653589793
//vector < vector<int>>a(0, vector<int>(0));二次元配列宣言


int main()
{
	int n,k;
	cin >> n>>k;
	vector<int>s(n+1);
	vector<int>a(n + 1);
	int i, j, b = 0;
	a.at(1) = 1;
	//long long int cnt = 0;
	for (i = 0; i < k; i++)
	{
		int l, r;
		cin >> l >> r;
		for (j = l; j <= r; j++)
		{
			s.at(b) = j;
			b++;
		}
	}
	sort(s.begin(),s.end());
	vector<int>t(b);
	for (i = 0; i < b; i++)
	{
		t.at(b-1-i) = s.at(n - i);
	}

	for (i = 2; i <= n; i++)
	{
		for (j = 0; j < b; j++)
		{
			
			//cnt++;
			if (i - t.at(j) >= 0)
			{
				a.at(i) = a.at(i) + a.at(i - t.at(j));
				a.at(i) = a.at(i) % 998244353;
			}
			else
			{
			//	cout << i << ',' << j << endl;
				break;
			}

		}
	}
		cout << a.at(n) << endl;
		//cout << cnt;
}