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>
#include <iostream>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define speed                     \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define ll long long
#define pb push_back
#define mem1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define endl "\n"
#define mod 1000000007
#define mod1 998244353
#define ff first
#define ss second
#define MAX 500005
#define N 10004
#define INF 1000000009
#define all(v) v.begin(), v.end()
#define sbit(a) __builtin_popcount(a)
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef pair<ll, ll> pll;
typedef pair<pll, ll> ppl;
typedef map<ll, ll> mpll;
typedef map<vector<ll>, ll> mpvl;
ll power(ll x, ll y, ll p)
{
    ll res = 1;
 
    x = x % p;
 
    if (x == 0)
        return 0;
    while (y > 0)
    {
 
        if (y & 1)
            res = (res * x) % p;
 
        y = y >> 1;
        x = (x * x) % p;
    }
 
    return res;
}
void solve()
{
 
    ll n, x, m;
    cin >> n >> x >> m;
    ll ans = x;
    ll t = x * x;
    ll c = 0;
    map<ll, ll> mp;
    n--;
    ll s[m + 1];
    mem0(s);
    ll tans = 0;
    ll tmp;
    for (ll i = 1; i <= (n,m + 1); i++)
    {
        ll md = t % m;
        if (!mp.count(md))
        {
            tans += md;
            t = md * md;
        }
        else
        {
            c = i - mp[md];
            tmp = mp[md];
            break;
        }
        mp[md] = i;
        s[i] = tans;
    }
    if (mp.count(0))
    {
        ans += tans;
        cout << ans << endl;
        return;
    }
    tmp--;
    n -= tmp;
    ll div = n / c;
    ll md = n % c;
    ans += (div * (s[c + tmp] - s[tmp]));
    ans += s[tmp];
    ans += s[tmp + md] - s[tmp];
    cout << ans << endl;
}
int main()
{
    speed;
    ll kk;
    kk = 1;
    //cin >> kk;
    while (kk--)
    {
 
        solve();
    }
 
    return 0;
}