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 <iomanip>
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define mod 998244353
#define vec vector<long long int>
#define fi first
#define se second
#define pb push_back
int main()
{
    ll n, k;
    cin >> n >> k;
    vector<bool> x(n + 1, 0);
    //vector<ll> y;
    vector<pair<ll, ll>> y;
    for (int i = 0; i < k; i++)
    {
        ll a, b;
        cin >> a >> b;
        pair<ll, ll> t(a, b);
        y.pb(t);
    }
    sort(y.begin(), y.end());

    ll m = 0;
    vector<bool> w(n + 1, 0);

    vector<ll> dp(n + 1, 0);
    dp[0] = 1;
    for (int i = 1; i < n + 1; i++)
    {
        for (int j = 0; j < y.size(); j++)
        {
            ll a = y[j].first;
            ll b = y[j].second;
            for (int k = a; k <= b; k++)
            {
                if (i - k >= 0)
                {
                    dp[i] = ((dp[i] % mod) + (dp[i - k]) % mod) % mod;
                }
            }
        }
    }
    cout << dp[n - 1];
}