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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll dp[12][200005];
void inp(ll a[],ll n)
{
    for(int i=0;i<n;i++)
    cin>>a[i];
}
void out(ll a[],ll n)
{
    for(int i=0;i<n;i++)
    cout<<a[i]<<" ";
    cout<<endl;
}
int z=998244353;
bool cmp(pair <int,int> f,pair <int,int> s)
{
  return (f.first<s.first);
}
void solve()
{
    ll n,k;
    cin>>n>>k;
    ll ans=0;
    vector <pair <ll,ll>> ranges;
    pair <ll,ll> p;
    //ranges.push_back(p);
    for(int i=1;i<=k;i++)
    {
        cin>>p.first>>p.second;
        ranges.push_back(p);
    }
    sort(ranges.begin(),ranges.end(),cmp);
    for(int i=0;i<=n;i++)
      for(int j=0;j<k+1;j++)
          dp[j][i]=0;
    for(int i=0;i<k+1;i++)
       dp[i][n]=0;
    dp[0][n]=1;
    for(int i=n-1;i>=1;i--)
    {
        for(int j=0;j<k;j++)
        {
            ll left=ranges[j].first;
            ll right=ranges[j].second;
            if(i+left<=n)
            {
              if((i+left)!=n)
              {dp[j][i]=dp[j][i+1];
               for(int l=0;l<k;l++)
               {
                   dp[j][i]=(dp[j][i]+dp[l][i+left])%z;
               }
              }
              else dp[j][i]=1;
            }
            else break;
            if(i+right<n)
            {
                for(int l=0;l<k;l++)
               {
                   dp[j][i]=(dp[j][i]-dp[l][i+right+1]+z)%z;
               }
            }
            else break;
        }
    }
    ll sum=0;
    for(int i=0;i<k;i++)
        sum=(sum+dp[i][1])%z;
    cout<<sum<<endl;
   /* for(int i=1;i<=n;i++)
        cout<<dp[0][i]<<" ";
    cout<<endl;
    for(int i=1;i<=n;i++)
        cout<<dp[1][i]<<" ";*/
    cout<<endl;
}
int main() {
    ll t=1;
    //cin>>t;
    while(t--)
    {
       solve();
    }
	return 0;
}