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>
using namespace std;
typedef long long ll;
#define all(x) (x).begin(), (x).end()
#define make_unique(x) \
    sort(all((x)));    \
    (x).resize(unique(all((x))) - (x).begin())
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1)
{
    cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args)
{
    const char *comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << " = " << arg1 << " | ";
    __f(comma + 1, args...);
}

ll max(ll a, ll b) { return (a > b ? a : b); }
ll min(ll a, ll b) { return (a < b ? a : b); }
ll dx[4] = {1, 0, -1, 0};
ll dy[4] = {0, 1, 0, -1};
void yes_or_no(ll x)
{
    if (x)
        cout << "Yes\n";
    else
        cout << "No\n";
}
const ll mod = 998244353;

void solve_test()
{
    ll ans = 0;
    string s;
    ll n, k;
    cin >> n >> k;
    vector<pair<ll, ll>> seg(k);
    for (int i = 0; i < k; i++)
        cin >> seg[i].first >> seg[i].second;
    vector<ll> a;
    for (auto i : seg)
    {
        a.push_back(i.first);
        a.push_back(i.second);
        trace(i.first, i.second);
    }
    set<ll> st;
    for (auto i : a)
        st.insert(i);
    ll mx = -1;
    for (ll i : a)
        mx = max(mx, i);

    //for (int i = 0; i < a.size(); i++)    cout << a[i] << " ";
    //   for (auto i : st) trace(i);
    vector<ll> dp(n + 1, 0);
    /*dp[0] = 1;
    for (auto j : st)
        for (ll i = j; i <= n; i++)
        {
            dp[i] = (dp[i] + dp[i - j]) % mod;
        }
        dp[0] = 1;
    */
    dp[1] = 1;
    for (ll i = 1; i <= n; i++)
    {
        for (auto j : st)
        {
            if (i - j >= 0)
                dp[i] = (dp[i] + dp[i - j]) % mod;
        }
    }
    ans = dp[n];
    cout << ans << "\n";
}
int main()
{
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    // cin >> t; while (t--)
    solve_test();
    return 0;
}