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;

using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
using vpi = vector<pii>;

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
    cout.tie(0);

    ll mod = 998244353;
    ll n, k;
    cin >> n >> k;
    set<ll> st;
    for (ll i = 0; i < k; i++) {
        ll x, y;
        cin >> x >> y;
        st.insert(x);
        st.insert(y);
    }
    vector<ll> a;
    for (ll x : st) a.push_back(x);
    sort(a.begin(), a.end());
    vector<ll> res(n, 0);
    res[0] = 1;
    for (ll j = 1; j < n; j++) {
        for (ll x : st) {
            if (j-x>=0) res[j] = (res[j] + res[j-x]) % mod;
        }
    }
    cout << res[n-1] << "\n";

	return 0;
}