Problem: 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.

Examples

Input

5 2
1 1
3 4


Output

4


Input

5 2
3 3
5 5


Output

0


Input

5 1
1 2


Output

5


Input

60 3
5 8
1 3
10 15


Output

221823067

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll mod = 998244353;

vector<int> L, R;
vector<ll> dp;

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);

	int n, k; cin >> n >> k; dp.resize(n + 1);
	L.resize(k); R.resize(k); dp[1] = 1; dp[0] = 0;
	for (int i = 0; i < k; i++) cin >> L[i] >> R[i];

	for (int i = 2; i <= n; i++) {
		dp[i] = dp[i - 1]; for (int j = 0; j < k; j++) if(i - L[j] > 0) 
			dp[i] = (dp[i] + dp[i - L[j]] - dp[max(0, i - R[j] - 1)] + mod) % mod;
	}

	cout << (dp[n] - dp[n - 1] + mod) % mod << '\n';
	cin.ignore(2); return 0;
}