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 <bits/stdc++.h>
#include <array>
#define int long long
using namespace std;
const int mod = 998244353;
const int N = 2e5 + 1;
struct node {
  int l, r, sum = 0, add = 0;
};
vector<node> t(4 * N);
void build(int l, int r, int v = 1) {
  t[v].l = l;
  t[v].r = r;
  if (l == r)
    return;
  int m = (r + l) >> 1;
  build(l, m, 2 * v);
  build(m + 1, r, 2 * v + 1);
}
void push(int v) {
  if (t[v].add == 0) return;
  if (t[v].l != t[v].r) {
    t[2 * v].add += t[v].add;
    t[2 * v + 1].add += t[v].add;
  }
  t[v].sum = (t[v].sum + t[v].add) % mod;
  t[v].add = 0;
}
void upd(int l, int r, int x, int v = 1) {
  push(v);
  if (l > r)
    return;
  if (t[v].l > r || t[v].r < l)
    return;
  if (l <= t[v].l && t[v].r <= r) {
    t[v].add = (t[v].add + x) % mod;
    return;
  }
  upd(l, r, x, 2 * v);
  upd(l, r, x, 2 * v + 1);
  t[v].sum = (t[2 * v].sum + t[2 * v + 1].sum) % mod;
}
int get(int pos, int v = 1) {
  push(v);
  if (pos > t[v].r || pos < t[v].l)
    return 0;
  if (t[v].l == t[v].r && t[v].l == pos)
    return t[v].sum % mod;
  return (get(pos, 2 * v) + get(pos, 2 * v + 1)) % mod;
}
signed main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n, k;
  cin >> n >> k;
  build(1, n);
  upd(1, 1, +1);
  vector<array<int, 2>> seg(k);
  for (int i = 0; i < k; ++i)
    cin >> seg[i][0] >> seg[i][1];
  for (int pos = 1; pos <= n - 1; ++pos) {
    for (int s = 0; s < k; ++s) {
      upd(seg[s][0] + pos, min(n, seg[s][1] + pos), get(pos));
    }
  }
  cout << get(n);
  return 0;
}