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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll mod = 998244353;
class SegmentTree {
private:
    int n;
    vector<ll> node, lazy;
    vector<bool> flag_lazy;
public:
    SegmentTree(int N) {
        n = 1;
        while (n < N) n *= 2;
        node.resize(2 * n - 1, 0);
        lazy.resize(2 * n - 1, 0);
        flag_lazy.resize(2 * n - 1, false);
    }

    void update(int i, int l, int r) {
        if (!flag_lazy[i]) return;
        node[i] += lazy[i];
        node[i] %= mod;
        if (r - l > 1) {
            lazy[2 * i + 1] += lazy[i] / 2;
            lazy[2 * i + 2] += lazy[i] / 2;
            flag_lazy[2 * i + 1] = true;
            flag_lazy[2 * i + 2] = true;
        }
        lazy[i] = 0;
    }

    void add(int a, int b, ll x, int k = 0, int l = 0, int r = -1) {
        if (r < 0) r = n;
        update(k, l, r);
        if (b <= l || r <= a) return;
        
        if (a <= l && r <= b) {
            lazy[k] += ((r - l) * (x));
            flag_lazy[k] = true;
            update(k, l, r);
        } else {
            add(a, b, x, 2 * k + 1, l, (l + r) / 2);
            add(a, b, x, 2 * k + 2, (l + r) / 2, r);
            node[k] = (node[2 * k + 1] + node[2 * k + 2]) % mod;
        }
    }

    ll query(int a, int b, int k = 0, int l = 0, int r = -1) {
        if (r < 0) r = n;
        update(k, l, r);
        if (b <= l || r <= a) return 0;        
        if (a <= l && r <= b) return node[k] % mod;

        ll sum_l = query(a, b, 2 * k + 1, l, (l + r) / 2);
        ll sum_r = query(a, b, 2 * k + 2, (l + r) / 2, r);
        return (sum_l + sum_r) % mod;
    }

    void disp() {
        for (int i : node) {
            cout << i << " ";
        }
        cout << "\n";
    }

};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, K;
    cin >> N >> K;
    int L[K], R[K];
    for (int i = 0; i < K; i++) {
        cin >> L[i] >> R[i];
    }

    SegmentTree st(500050);
    st.add(1, 2, 1);
    
    for (int i = 1; i < N; i++) {
        ll x = st.query(i, i + 1);
        for (int j = 0; j < K; j++) {
            int l = i + L[j];
            int r = i + R[j] + 1;
            
            st.add(min(l, N + 1), min(r, N + 1), x);
            //st.add(l, r, x);
        }
        //cout << x << "\n";
    }

    cout << st.query(N, N + 1) << "\n";

    return 0;
}