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 y_n(ans) cout<<(ans?"Yes":"No")<<endl;

const long long mod = 998244353;
const long long check_m = 1e9 + 7;
const int MAX_N = 5 * 1e5 + 100;
const int MAX_n = 3 * 1e5;

int main() {
    long long n, k;
    cin >> n >> k;
    vector<long long> l(k), r(k);
    for (int i = 0; i < k; ++i) {
        cin >> l[i] >> r[i];
        r[i]++;
    }
    vector<long long> num(MAX_N, 0), chnum(MAX_N, 0);
    num[1] = 1;
    num[2] = -1;
    chnum[1] = 1;
    chnum[2] = -1;
    vector<long long> ans(MAX_n, 0), check(MAX_n, 0);
    ans[1] = num[1];
    check[1] = num[1];
    vector<long long> mod_times(MAX_N, 0);
    for (int i = 1; i < MAX_n; ++i) {
        if (i != 1) {
            ans[i] = ans[i-1] + num[i];
            ans[i] %= mod;
            while (ans[i] < 0) ans[i] += mod;

            check[i] = check[i-1] + chnum[i];
            check[i] %= check_m;
            while (check[i] < 0) check[i] += check_m;
        }

        if (ans[i] == 0 && check[i] == 0) {
            continue;
        }
        for (int j = 0; j < k; ++j) {
            num[i + l[j]] += ans[i];
            num[i + r[j]] -= ans[i];
            chnum[i + l[j]] += check[i];
            chnum[i + r[j]] -= check[i];
        }
    }
    cout << ans[n] << endl;
}