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 <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const long double pi=acos(-1);
#define x  first
#define y  second
#define read(n,a) for(int i=0;i<n;i++)cin>>a[i];
#define rep(i,z,n)for(int i=z;i<n;i++)
#define all(v)    v.begin(),v.end()

//int a[200002]={};
//int b[200002]={};
vector<int>v;
ll n,m;
ll MOD=998244353;
ll dp[200002];
ll sol(ll cu)
{
    if(cu==0)return 1;
    ll &ans=dp[cu];
    if(ans!=-1)return ans;
    ans=0;
    for(int i=0;i<v.size();i++)
    {
        if(v[i]>cu)break;
        else ans=(ans%MOD+sol(cu-v[i])%MOD)%MOD;
    }
    return ans%MOD;
}
int main()
{
    cin.tie(0);
    cin.sync_with_stdio(0);
    ///freopen("ss.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int t=1;
    //cin>>t;
    while(t--)
    {
        cin>>n>>m;
        pair<int,int>a[m];
        rep(i,0,m)
        {
            cin>>a[i].x>>a[i].y;
            if(a[i].x>a[i].y)swap(a[i].x,a[i].y);
        }
        rep(i,0,m)
        {
            for(int j=a[i].x;j<=a[i].y;j++)v.push_back(j);
        }
        sort(all(v));
        ll d=n-1;
        rep(i,0,200002)dp[i]=-1;
        cout<<sol(d);
    }
    return 0;
}