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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <functional>
#include <unordered_map>
#include <array>

using namespace std;

#define ll long long
#define all(v) v.begin(),v.end()
#define bootstrap ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);

// custom hash can be a standalone function object:
struct MyHash {
    std::size_t operator()(pair<long, long> const &p) const noexcept {
        std::size_t h1 = std::hash<long>{}(p.first);
        std::size_t h2 = std::hash<long>{}(p.second);
        return h1 ^ (h2 << 1); // or use boost::hash_combine (see Discussion)
    }
};

struct MyComp {
    template<typename T>
    std::size_t operator()(T &l, T &r) const noexcept {
        return l < r;
    }
};

void solve() {
    int n, k;
    cin >> n >> k;
    vector<pair<int, int>> corners;
    int l, r;
    for (int i = 0; i < k; i++) {
        cin >> l >> r;
        corners.push_back(pair<int, int>(l, r));
    }
    sort(corners.begin(), corners.end());

    long CON = 998244353;
    array<long, 2 * 100000 + 1> dp;
    dp[1] = 1;
    for (int i = 1; i <= n; i++) {
        for (auto &corner: corners) {
            if( i + corner.first > n) {
                break;
            }
            for (int j = corner.first; j <= corner.second; ++j) {
                if( i + j > n) {
                    break;
                }
                dp[i + j] += dp[i];
                dp[i + j] %= CON;
            }
        }
    }
    cout << dp[n];
}

int main() {
    bootstrap
    solve();
}