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)

/*
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@TT_beginner
*/


#include "bits/stdc++.h"
using namespace std;
#define ok1 cerr<<"ok1\n";
#define ok2 cerr<<"ok2\n";
#define M LLONG_MAX
#define rep(i,n) for(ll i=0;i<n;++i)
#define REP(i,s,n) for(ll i=(s);i<(n);++i)
#define repr(i,n) for(ll i=n-1;i>=0;--i)
#define REPR(i,s,n) for(ll i=(s);i>=(n);--(i))
#define all(a) (a).begin(),(a).end()
#define reall(a) (a).rbegin(),(a).rend()
#define pb emplace_back //emplace_backの方が速いが使い慣れてないため
#define DOUBLE fixed << setprecision(15)
#define fi first
#define se second
#define mp make_pair
#define mt make_tuple
#define out(x, y) ((x) < 0 || h <= (x) || (y) < 0 || w <= (y))
#define TO_STRING(VariableName) # VariableName
#define debug(x) cerr<<TO_STRING(x)<<" is: "<<x<<" "<<endl;
const double pi = acos(-1.0L);
typedef vector<int> vi;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<char> vc;
typedef vector<double> vd;
typedef vector<bool> vb;
typedef deque<ll> dll;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> TLL;
typedef vector<P> vP;
typedef vector<TLL> vTLL;
//const ll mod = 1e9 + 7;
const ll mod = 998244353;
ll dx[8] = { -1,0,1,0,1,1,-1,-1 };
ll dy[8] = { 0,-1,0,1,1,-1,1,-1 };
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>&p) {
	ost << "{ " << p.first << ", " << p.second << " }";
	return ost;
}
template<class T>
ostream& operator<<(ostream& ost, const vector<T>&v) {
	ost << "{ ";
	for (int i = 0; i<v.size(); i++) {
		if (i)ost << ", ";
		ost << v[i];
	}
	ost << " }";
	return ost;
}
template<class A, class B>
ostream& operator<<(ostream& ost, const map<A, B>&v) {
	ost << "{ ";
	for (auto p : v) {
		ost << "{ " << p.first << ", " << p.second << " }";
	}
	ost << " }";
	return ost;
}
void debugs() { std::cerr << "\n"; }
template <class T, class... Args>
void debugs(const T &x, const Args &... args) {
	std::cerr << x << " ";
	debugs(args...);
}
bool out_check(ll a, ll b) { return (0 <= a && a < b); }
double pitagoras(ll a, ll b, ll c, ll d) {
	double dx = a - b, dy = c - d;
	return pow(dx * dx + dy * dy, 0.5);
}
ll modpow(ll a, ll b, ll mo = mod) { ll c = 1; while (b > 0) { if (b & 1) { c = a * c%mo; }a = a * a%mo; b >>= 1; }return c; }
void Yes(bool x) { cout << ((x) ? "Yes\n" : "No\n"); }
void YES(bool x) { cout << ((x) ? "YES\n" : "NO\n"); }
void yes(bool x) { cout << ((x) ? "yes\n" : "no\n"); }
void Yay(bool x) { cout << ((x) ? "Yay!\n" : ":(\n"); }
void OK(bool x) { cout << ((x) ? "OK!\n" : "NG\n"); }
//*
ll  n, m, d, e, l, r, k, h, w, Q, ans, ret = M;
bool flag = false, flag2 = false, flag3 = false;
string s, t, u;
//*

/* For AtCoder Only
#include "atcoder/all"
using namespace atcoder;
//*/

int main() {
	//cin.tie(0); ios::sync_with_stdio(false);//not interactive
	cin >> n >> k;
	vP v;
	rep(i, k) {
		cin >> l >> r;
		v.push_back(P(l, r+1));
	}
	sort(all(v));
	vll dp(n);
	dp[0] = 1;
	
	rep(i, n) {
		for (auto& p : v) {
			REP(d, p.first, p.second) {
				if (i - d < 0)continue;
				dp[i] += dp[i - d];
				dp[i] %= mod;
			}
		}
	}
	cout << dp[n - 1]%mod << endl;
}