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 fi first
#define se second
#define maxn 200003
#define mod 998244353
#define pb push_back
using namespace std;
typedef long long LL;
typedef pair<int,int> pi;

int n,k;
int dp[maxn];
int sum[maxn];
int l[10],r[10];

int main() {
	scanf("%d%d",&n,&k);
	for( int i = 0 ; i < k ; i++ )
		scanf("%d%d",&l[i],&r[i]);
	dp[n] = 1;
	sum[n] = 1;
	for( int i = n-1 ; i >= 1 ; i-- ) {
		for( int j = 0 ; j < k ; j++ ) {
			int ll = min(i+l[j],n+1);
			int rr = min(i+r[j],n+1);
			int add = (sum[ll] - sum[rr+1] + mod) % mod;
			dp[i] = (dp[i] + add) % mod;
		}
		sum[i] = (sum[i+1] + dp[i]) % mod;
	}
	printf("%d\n",dp[1]);
	return 0;
}