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>
#define rep(i,n) for(int i=0;i<(int)n;++i)
#define REP(i,a,b) for(int i=int(a);i<(int)b;++i)
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<ll>
#define vvl vector<vecctor<ll>>
#define vp vector<pair<ll,ll>>
#define pb push_back
#define all(v) v.begin(),v.end()



using namespace std;
using ll = long long;
const ll INF = 1e18;
const double PI = acos(-1);

template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

const ll mod = 998244353;



int main(void)
{
    std::cin.tie(nullptr);
    std::ios::sync_with_stdio(false);
    ll n,k;
    cin >> n >> k;
    vl l(n),r(n);
    rep(i,k){
        cin >> l[i] >> r[i];
    }

    ll dp[200000 + 5];
    ll dp_sum[200000 + 5];
    dp[0] = 1;
    dp_sum[0] = 0;
    dp_sum[1] = 1;
    for(int i = 1;i < n; ++i){
        rep(j,k){
            int ri = i-l[j],li = i-r[j];
            if(ri < 0)continue;
            li = max(0,li);
            dp[i] += dp_sum[ri+1]-dp_sum[li];
            dp[i] %= mod;
        }
        dp_sum[i+1] = dp_sum[i] + dp[i];
        dp_sum[i+1] %= mod;
    }
    ll ans = dp[n-1];
    cout << (ans%mod + mod)%mod << endl;

    return 0;
}