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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_imports)]
use proconio::input;
use std::cmp::max;
use std::cmp::min;
use std::collections::HashSet;
// use std::collections::VecDeque;

fn main() {
    input! {
        N: usize,
        K: usize,
        LRs: [(usize, usize); K],
    }

    const MOD: i64 = 998244353;

    // let mut dp = vec![0i64; N];
    // dp[0] = 1;

    // acc[0] := 0
    // acc[i+1] := acc[i] + dp[i]
    let mut acc = vec![0i64; N + 1];
    acc[0] = 0;
    acc[1] = 1; // dp[0]

    for n in 1..N {
        let mut cand: i64 = 0;

        for &(l, r) in LRs.iter() {
            if n < l {
                continue;
            }

            // === 累積和で表現しないver ===
            // let mut acc : i64 = 0;
            // for i in l..=min(r,n) {
            //     // MODを消してたら累積和に気付けたかも?
            //     // acc += dp[n - i];
            //     acc = (acc + dp[n - i]) % MOD;
            // }
            // cand = (cand + acc) % MOD;
            // === ここまで ===

            // dp[n-r] + dp[n-(r-1)] + ... dp[n-l]
            cand = (cand + acc[n - l + 1] - acc[n - min(r, n)] + MOD) % MOD;
        }

        // dp[n] = cand;
        acc[n + 1] = (acc[n] + cand) % MOD;
    }

    let ans = (acc[N] - acc[N - 1] + MOD) % MOD;
    println!("{}", ans);
}