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;

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep3(i,m,n) for(int (i)=m;(i)<=(n);(i)++)
#define rep3rev(i,m,n) for(int (i)=m;(i)>=(n);(i)--)
#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define eb emplace_back

using ll = long long;
using vll = vector<ll>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using P = pair<int, int>;

const int MOD = 998244353;

// Binary Indexed Tree (Fenwick Tree)
// https://youtu.be/lyHk98daDJo?t=7960
template<typename T>
struct BIT {
    int n;
    vector<T> d;
    BIT(int n=0):n(n),d(n+1) {}
    void add(int i, T x=1) {
        for (i++; i <= n; i += i&-i) {
            d[i] += x;
            d[i] %= MOD;
        }
    }
    T sum(int i) {
        T x = 0;
        for (i++; i; i -= i&-i) {
            x += d[i];
            x %= MOD;
        }
        return x;
    }
    T sum(int l, int r) {
        return (sum(r-1) + MOD - sum(l-1)) % MOD;
    }
};

void Main(){
    int n, k; cin >> n >> k;
    vi l(k), r(k);
    rep(i, k){
        cin >> l[i] >> r[i];
    }

    BIT<int> dp(n+5);
    dp.add(1, 1);
    
    rep3(i, 2, n){
        rep(j, k){
            if(i - l[j] <= 0) continue;
            int L = max(0, i-r[j]);
            int R = i-l[j];
            int tmp = dp.sum(L, R+1);
            // cout << i << " " << j << " " << tmp << endl;
            dp.add(i, tmp);
        }
    }
    cout << dp.sum(n, n+1) << endl;
    return;
}

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
    return 0;
}