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 rep(i,n) for(int i = 0; i < (n); i++)
#define rrep(i,n) for(int i = (n)-1; i >= 0; i--)
#define rep1(i,n) for(int i = 1; i <= (n); i++)
#define rrep1(i,n) for(int i = (n); i > 0; i--)

#define ll long long
#define pi pair<int, int>
#define pll pair<ll, ll>

#define MOD 998244353LL
#define INF 1000000000000000LL

using namespace std;

// 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] = (d[i]+x)%MOD;
    }
  }
  T sum(int i) {
    T x = 0;
    for (i++; i; i -= i&-i) {
      x = (x+d[i])%MOD;
    }
    return x%MOD;
  }
};

int main(){
  int n,k;cin>>n>>k;
  vector<pi>lr(k);
  rep(i, k)cin>>lr[i].first>>lr[i].second;

  BIT<ll> bit(n+2);
  bit.add(1, 1);
  bit.add(2, -1);

  rep1(i, n){
    ll cv = bit.sum(i)%MOD;
    rep(j, k){
      int l=lr[j].first;
      int r = lr[j].second;
      l = i+l;
      r = i+r+1;

      if(l>n)continue;
      if(r>n)r=n+1;

      bit.add(l, cv);
      bit.add(r, -cv);
    }
  }

  cout<<bit.sum(n)%MOD<<endl;

  return 0;
}