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 mod 998244353
using namespace std;
typedef long long int ll;
typedef long double ld;
const int maxn=2E5+5;
ll n,k,l[maxn],r[maxn],f[maxn];
int main()
{
	ios::sync_with_stdio(false);
	cin>>n>>k;
	for(int i=1;i<=k;++i)
		cin>>l[i]>>r[i];
	f[1]=1;
	ll s=0;
	for(int i=2;i<=n;++i)
	{
		for(int j=1;j<=k;++j)
		{
			if(i-l[j]>=1)
				s=(s+f[i-l[j]])%mod;
			if(i-r[j]-1>=1)
				s=(s-f[i-r[j]-1])%mod;
		}
		f[i]=s;
	}
	cout<<(f[n]%mod+mod)%mod<<endl;
	return 0;
}