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: Wrong Answer)

use proconio::input;

#[allow(non_snake_case)]
fn main() {
    const MOD: i32 = 998244353;
    input!{
        N: usize,
        K: usize,
        LR: [(usize, usize); K]
    }
    let mut dp = vec![0; N];
    let mut a = vec![0; N];
    for (l, r) in LR.iter() {
        if *l < N {
        a[*l] = 1;
        }
        if r + 1 < N {
        a[r+1] = -1;
        }
    }
    for i in 1..N {
        for (l, r) in LR.iter() {
            dp[i] = (dp[i-1] + a[i]) % MOD;
            if i + l < N {
            a[i+l] += dp[i];
            a[i+l] %= MOD;
            }
            if i + r + 1 < N {
            a[i+r+1] -= dp[i];
            a[i+r+1] %= MOD;
            }
        }
    }
    let mut ans = dp[N-1];
    if ans < 0 {
        ans += MOD;
    }
    println!("{}", ans);
}