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 pair<int,int> pii;
#define int long long int //**
#define pb push_back
#define mp make_pair

vector<int> BIT;

const int mod = 998244353;

void update(int pos,int val)
{
    for(int i=pos;i<=BIT.size()-1;i+=(i&(-i)))
    {
        // BIT[i] = (BIT[i]%mod + val%mod)%mod;
        BIT[i] = BIT[i] + val;
    }
}

int getSum(int pos)
{
    int ret = 0;
    for(int i=pos;i>0;i-=(i&(-i)))
    {
        // ret = (ret%mod + BIT[i]%mod)%mod;
        ret = ret + BIT[i];
    }
    return ret;
}

void solve()
{
    int n,k;cin >> n >> k;
    vector<pii> ranges;
    for(int i=0;i<k;++i)
    {
        int l,r; cin >> l >> r;
        ranges.pb(mp(l,r));
    }

    BIT = vector<int>(n+1);

    update(1,1);
    update(2,-1);

    /*
    cout << " THE INITIAL VALUE IS : " << ' ' << getSum(1) << '\n';
    */

    for(int i=1;i<=n;i++)
    {

        if(getSum(i))
        {
            int val = getSum(i);
            // cout << " THIS IS FROM WHERE I AM GOING TO PUSH MY VALUES : " << i << " AND THE VALUE I AM CONTRIBUTING IS : " << val << '\n';
            for(int j=0;j<(int)ranges.size();++j)
            {
                int l = ranges[j].first, r = ranges[j].second;

                update(i+l,val);
                update(i+r + 1,-1* val);
            }
        }
    }
    /*
    for(int i=1;i<=n;++i)
    {
        cout << " FOR THE : " << i << " CITY : " << " THE VALUES IS : " << getSum(i) << '\n';
    }
    */
    
    int ans = getSum(n);
    cout << ans%mod << '\n';

}
 
signed main()
{
     int test_cases;test_cases=1;
     //cin >> test_cases;
     while(test_cases --> 0) solve();
}