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;

#define PB push_back
#define f first
#define s second
#define what_is(x) cerr << #x << " is " << x << endl;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;

const int INF = 1000000007;
const int MOD = 998244353;

int bit[200005];
int n, k;
void upd(int x, int val)
{
    for(; x <= n; x += x&-x)
    {
        bit[x] = (bit[x] + val) % MOD;
    }
}

int get(int x)
{
    int res = 0;
    for(; x>0; x -= x&-x)
    {
        res = (res + bit[x]) % MOD;
    }
    return res;
}
void solve_test()
{
    cin >> n >> k;
    upd(1, 1);
    vector<pii> v;
    for(int i=0; i<k; i++)
    {
        int l, r;
        cin >> l >> r;
        v.PB({l, r});
    }
    sort(v.begin(), v.end());
    for(int i=2; i<=n; i++)
    {
        int lo = 0, hi = 0;
        for(int j=0; j<k; j++)
        {
            lo = max(lo, v[j].f);
            hi = v[j].s;
            if(lo > hi)
                continue;
            if(lo >= i)
                break;
//            what_is(lo);
//            what_is(hi);
            hi = min(i-1, hi);
            int res = (MOD + get(i - lo) - get(i - hi - 1)) % MOD;
            upd(i, res);
            lo = hi+1;
        }
       //cout << i << " " << get(i) << "\n";
    }
    cout << get(n) - get(n-1) << "\n";

}

signed main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    int tests;
    tests = 1;
    //cin >> tests;
    while(tests--)
    {
        solve_test();
    }
    return 0;
}