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)

/***********************************************************
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					huzaifa242©

***********************************************************/
#include "bits/stdc++.h"
using namespace std;
#define int int64_t
#define MAX 100005
#define x first 
#define y second

int seed;
mt19937 rng(seed=chrono::steady_clock::now().time_since_epoch().count());
inline int rnd(int l = 0, int r = numeric_limits <int>::max())
{return uniform_int_distribution<int> (l, r)(rng);}

template <typename T, typename G>
ostream& operator << (ostream& out, const pair <T, G> &a) 
{return out << "( " << a.x << ", " << a.y << ")";}

template <typename T>
ostream& operator << (ostream& out, const vector <T> &a) 
{out << "[ "; for(const auto &i: a) out << i << ", "; return out << "]" << endl;}

template <typename T>
ostream& operator << (ostream& out, const set <T> &a) 
{out << "{ "; for(const auto &i: a) out << i << ", "; return out << "}" << endl;}

template <typename T, typename G>
ostream& operator << (ostream& out, const map <T, G> &a) 
{out << "< "; for(const auto &i: a) out << i << ", "; return out << ">" << endl;}

template <typename T, size_t N>
typename enable_if<!is_same <typename remove_cv <T>::type, char>::value,
ostream&>::type operator << (ostream& out, T (&a)[N])
{out << "[ "; for(const auto &i: a) out << i << ", "; return out << "]" << endl;}

void debug_out(){cerr << endl;} 
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T)
{cerr << (H) << ' '; debug_out(T...);}
 
#ifdef LOCAL
#define debug(...) cerr << '#' << #__VA_ARGS__ << ":\n", debug_out(__VA_ARGS__)
#else
#define debug(...) 24
#endif

const int mod = 998244353;
int md(int x, int mdo = mod) {
	x %= mdo;
	return x < 0? x + mdo: x;
}

signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	int n, k;
	cin >> n >> k;
	vector <pair <int,int> > segs(k);
	for (auto &seg : segs) 
		cin >> seg.x >> seg.y;
	vector <int> count(n + 1, 0), sum(n + 1, 0);
	count[1] = sum[1] = 1;
	for (int i = 2; i <= n; ++i) {
		for (auto &seg : segs) {
			int l = max(i - i + 1, i - seg.y);
			int r = max(i - i + 1, i - seg.x);
			if (i - seg.x <= 0)
				continue; 
			count[i] = md(count[i] + sum[r] - sum[l - 1]);
		}
		sum[i] = md(sum[i - 1] + count[i]);
	}
	cout << count[n] << '\n';
	return 0;
}