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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define V vector
typedef long long ll;
typedef unsigned long long ull;

typedef V<int> vi;
typedef V<ll> vll;
typedef V<string> vs;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define f(i,a) for(int i = 0 ; i < a ; i++)
#define fll(i,a) for(ll i = 0 ; i < a ;i++)
#define forab(i,a,b) for(int i = a ; i < b ; i++)
#define prec(x) cout<<fixed<<setprecision(x)
#define ff first
#define ss second

#define pb push_back
#define mp make_pair
#define numberofdigits(x) floor(log10(x)) + 1
#define all(x) (x).begin() ,(x).end()

const ll mod = 998244353;

ll binpow(ll a, long long b) {
	ll res = 1;
	while (b > 0) {
		if (b & 1)
			res = (res * a);
		a = (a * a);
		b >>= 1;
	}
	return res;
}
const int N = 1e6 + 5;

void solve() {

	int n , k;
	cin >> n >> k;

	int dp[n + 1] , pref[n + 1];
	memset(dp , 0 , sizeof(dp));
	memset(pref , 0 , sizeof(pref));
	dp[1] = 1 , pref[1] = 1;

	int l[k] , r[k];
	for (int i = 0 ; i < k ; i++)
		cin >> l[i] >> r[i];

	for (int i = 2 ; i <= n ; i++) {

		for (int j = 0 ; j < k ; j++) {

			if (i <= l[j]) continue;
			int L = l[j];
			int R = min(r[j] , i - 1);

			dp[i] = (dp[i] % mod + (pref[i - L] - pref[i - R - 1]) % mod) % mod;
		}

		pref[i] = (pref[i - 1] % mod + dp[i] % mod) % mod;
	}

	cout << dp[n] << "\n";
}

int main() {

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);


	int t = 1;
	//cin >> t;

	while (t--)
		solve();
	return 0;
}