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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

fun main() {
    val (n, K) = readLine()!!.split(" ").map { it.toInt() }
    val s = Array(K) {
        readLine()!!.split(" ").map { it.toInt() }.let {
            Pair(it[0], it[1])
        }
    }

    val mod = 998244353L

    val dp = MutableList(n + 1) { 0L }
    val dpSum = MutableList(n + 1) { 0L }
    dp[1] = 1L

    for (i in 1 until n + 1) {
        for ((l, r) in s) {
            dp[i] = dp[i] % mod + ((if ((i - l) > 0) dpSum[i - l] else 0) % mod - (if ((i - r - 1) > 0) dpSum[i - r - 1] else 0) % mod) % mod
        }
        dpSum[i] = (dpSum[i - 1] + dp[i]) % mod
    }
    println(dp[n] % mod)
}