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++)
using namespace std;
using ll = long long;
using P = pair<int, int>;
using V = vector<int>;

int main(int argc, char const *argv[])
{
    int mod = 998244353;
    int n, k; cin >> n >> k;
    int l[15], r[15];
    for (int i = 0; i < k; i++) {
        cin >> l[i] >> r[i];
    }
    
    sort(l,l+k);
    sort(r,r+k);

    long long int dp[200010] = {};
    dp[1] = 1;
  	if (l[0] == 1) dp[2] = 1;
    for (int i = 3; i <= n; i++)
    {
        dp[i] = dp[i-1];
        for (int j = 0; j < k; j++)
        {
            if (i-l[j]-1 >= 0)
            {
                dp[i] += dp[i-l[j]];
                if (i-r[j]-1 >= 1) dp[i] -= dp[i-r[j]-1];
              dp[i]+= mod;
              dp[i]%= mod;
            }
            
        }
      cout << dp[i] << endl;
    }
    cout << dp[n] << endl;


    

    return 0;
}