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 <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < (n); i++)
using namespace std;
const int MOD = 998244353;

int main() {
	cin.tie(0)->sync_with_stdio(false);

	int n, k; cin >> n >> k;
	vector<int> L(k), R(k);
	REP(i, k) cin >> L[i] >> R[i];
	vector<int> dp(n + 1, 0), s(n + 1, 0);
	dp[1] = s[1] = 1;
	for(int i = 2; i <= n; i++) {
		REP(j, k) {
			int l = i - R[j];
			int r = i - L[j];
			if (r < 1) continue ;
			l = max(l, 1);
			int sum = s[r] - s[l-1];
			sum %= MOD;
			sum += MOD;
			sum %= MOD;
			dp[i] += sum;
			dp[i] %= MOD;
		}
		s[i] = s[i-1] + dp[i];
		s[i] %= MOD;
	}
	cout << dp[n] << '\n';
	return 0;
}