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 <iostream>
#include <algorithm>
#include <limits>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <vector>
#include <deque>
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <cmath>

using namespace std;
const int mod = 998244353;
int n, k;

int main() {
    scanf("%d %d", &n, &k);
    vector<pair<int, int>> s;
    for (int i = 0; i < k; ++i) {
        int l, r;
        scanf("%d %d", &l, &r);
        s.push_back({l, r});
    }


    vector<long long> dp(n, 0);
    vector<long long> preSum(n, 0);
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j < k; ++j) {
            int l = i - s[j].second;
            int r = i - s[j].first;
            if (r < 0) continue;
            l = max(1, l);
            dp[i] += preSum[r] - preSum[l - 1];
            dp[i] %= mod;
        }
    }
    printf("%lld\n", dp[n]);
    return 0;


    sort(s.begin(), s.end());
    // cnt[i] := the number of possible ways to reach ith cell
    // cnt[i] = sum{cnt[i-k] | k in S}
    vector<long long> cnt(n + 1, 0);
    cnt[1] = 1;
    for (int i = 2; i <= n; ++i) {
        for (auto it : s) {
            for (int k = it.first; k <= it.second; ++k) {
                if (i - k <= 0) {
                    break;
                }
                cnt[i] += cnt[i - k];
            }
            cnt[i] %= mod;
        }
    }
    printf("%lld\n", cnt[n]);
    return 0;
}