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>
#define ll long long
#define dbg(x) cout<<#x<<": "<<x<<endl;
#define N 300005
#define M 998244353
#define pii pair<ll,ll>
#define fast ios_base::sync_with_stdio(0);cin.tie(0);
using namespace std;
ll suf[N];
ll dp[N];

main()
{
    ll n,k;
    cin>>n>>k;
    vector<pii>v;
    v.resize(k);
    for(ll i=0;i<k;i++)
    {
        cin>>v[i].first>>v[i].second;
    }
    sort(v.begin(),v.end());

    dp[n]=1;
    suf[n]=1;

    for(ll i=n-1;i>=1;i--)
    {
        ll now=0;
        for(ll j=0;j<k;j++)
        {
            ll l=min(n,i+v[j].first),r=min(n,i+v[j].second);
            now+=suf[l]-suf[r+1];
            now%=M;
            now=(now+M)%M;
        }
        dp[i]=now;
        suf[i]=suf[i+1]+dp[i];
        suf[i]%=M;
    }
    cout<<dp[1];
}