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: Accepted)

#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;

ll mod = 998244353;

struct LazySegmentTree {
private:
    int n;
    vector<ll> node, lazy;

public:
    LazySegmentTree(vector<ll> v) {
        int sz = (int)v.size();
        n = 1; while(n < sz) n *= 2;
        node.resize(2*n-1);
        lazy.resize(2*n-1, 0);

        for(int i=0; i<sz; i++) node[i+n-1] = v[i];
        for(int i=n-2; i>=0; i--) node[i] = node[i*2+1] + node[i*2+2];
    }

    void eval(int k, int l, int r) {
        if(lazy[k] != 0) {
            node[k] += lazy[k];
            if(r - l > 1) {
                lazy[2*k+1] += lazy[k] / 2;
                lazy[2*k+2] += lazy[k] / 2;
            }
            lazy[k] = 0;
        }
    }

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

int main() {
  ll N, K; cin >> N >> K;
  vector<ll> L(K), R(K);
  for (ll i = 0; i < K; i++) {
    cin >> L[i] >> R[i];
    R[i]++;
  }
  LazySegmentTree seg( vector<ll>(N, 0) );
  seg.add(0, 1, 1);
  for (ll i = 0; i < N; i++) {
    ll tgt = seg.getsum(i, i+1);
    if (tgt == 0) continue;
    for (ll j = 0; j < K; j++) {
      ll left = i+L[j];
      ll right = i+R[j];
      if (left < N) {
        if (right > N) right = N;
        seg.add(left, right, tgt);
      }
    }
  }
  cout << seg.getsum(N-1, N) << endl;
  return 0;
}