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)

use proconio::input;

const MOD: usize = 998244353;

fn main() {
    input!{
        n: usize,
        k: usize,
        lr: [(usize, usize); k],
    }
    let mut dp = vec![0; n+1];
    let mut diff = vec![0; n+1];
    dp[0] = 1;
//    dp[1] = MOD-1;
    for i in 0..n {
        if i > 0 {
            dp[i] += diff[i];
            dp[i] %= MOD;
        }
        for (l, r) in lr.iter() {
            if i + l < n {
                diff[i+l] += dp[i];
                diff[i+l] %= MOD;
            }
            if i + r + 1 < n {
                diff[i+r+1] = diff[i+r+1] + MOD - dp[i];
                diff[i+r+1] %= MOD;
            }
        }
        diff[i+1] += diff[i];
        diff[i+1] %= MOD;
    }
    println!("{}", dp[n-1]);
}