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 ll long long
#define F first
#define S second
const int A = 2e5 + 2;
const int mod = 998244353;
int add(ll x, ll y) {
    return (x + y + mod) % mod;
}
int n, k;
int l[11], r[11];
int dp[A], freq[A];
int solve() {
    dp[1] = freq[1] = 1;
    for (int i = 2; i <= n; i++) {
        for (int j = 0; j < k; j++) {
            int up = max(0, i - l[j]), low = max(1, i - r[j]);
            dp[i] = add(dp[i], add(freq[up], -freq[low - 1]));
        }
        freq[i] = add(dp[i], freq[i - 1]);
//        cout << dp[i] << '\n';
    }
    return dp[n];
}
int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> k;
    for (int i = 0; i < k; i++)
        cin >> l[i] >> r[i];
    cout << solve();
    return 0;
}