An ant moves on the real line with constant speed of 1 unit per second. It starts at 0 and always moves to the right (so its position increases by 1 each second).

There are n portals, the i-th of which is located at position x_i and teleports to position y_i < x_i. Each portal can be either active or inactive. The initial state of the i-th portal is determined by s_i: if s_i=0 then the i-th portal is initially inactive, if s_i=1 then the i-th portal is initially active. When the ant travels through a portal (i.e., when its position coincides with the position of a portal): 

  * if the portal is inactive, it becomes active (in this case the path of the ant is not affected); 
  * if the portal is active, it becomes inactive and the ant is instantly teleported to the position y_i, where it keeps on moving as normal. 



How long (from the instant it starts moving) does it take for the ant to reach the position x_n + 1? It can be shown that this happens in a finite amount of time. Since the answer may be very large, compute it modulo 998 244 353.

Input

The first line contains the integer n (1≤ n≤ 2⋅ 10^5) — the number of portals.

The i-th of the next n lines contains three integers x_i, y_i and s_i (1≤ y_i < x_i≤ 10^9, s_i∈\{0,1\}) — the position of the i-th portal, the position where the ant is teleported when it travels through the i-th portal (if it is active), and the initial state of the i-th portal.

The positions of the portals are strictly increasing, that is x_1<x_2<⋅⋅⋅<x_n. It is guaranteed that the 2n integers x_1,   x_2,   ...,   x_n,   y_1,   y_2,   ...,   y_n are all distinct.

Output

Output the amount of time elapsed, in seconds, from the instant the ant starts moving to the instant it reaches the position x_n+1. Since the answer may be very large, output it modulo 998 244 353.

Examples

Input


4
3 2 0
6 5 1
7 4 0
8 1 1


Output


23


Input


1
454971987 406874902 1


Output


503069073


Input


5
243385510 42245605 0
644426565 574769163 0
708622105 208990040 0
786625660 616437691 0
899754846 382774619 0


Output


899754847


Input


5
200000000 100000000 1
600000000 400000000 0
800000000 300000000 0
900000000 700000000 1
1000000000 500000000 0


Output


3511295

Note

Explanation of the first sample: The ant moves as follows (a curvy arrow denotes a teleporting, a straight arrow denotes normal movement with speed 1 and the time spent during the movement is written above the arrow). $$$ 0 \stackrel{6}{\longrightarrow} 6 \leadsto 5 \stackrel{3}{\longrightarrow} 8 \leadsto 1 \stackrel{2}{\longrightarrow} 3 \leadsto 2 \stackrel{4}{\longrightarrow} 6 \leadsto 5 \stackrel{2}{\longrightarrow} 7 \leadsto 4 \stackrel{2}{\longrightarrow} 6 \leadsto 5 \stackrel{4}{\longrightarrow} 9  Notice that the total time is 6+3+2+4+2+2+4=23$$$.

Explanation of the second sample: The ant moves as follows (a curvy arrow denotes a teleporting, a straight arrow denotes normal movement with speed 1 and the time spent during the movement is written above the arrow). $$$ 0 \stackrel{454971987}{\longrightarrow} 454971987 \leadsto 406874902 \stackrel{48097086}{\longrightarrow} 454971988  Notice that the total time is 454971987+48097086=503069073$$$.

Explanation of the third sample: Since all portals are initially off, the ant will not be teleported and will go straight from 0 to x_n+1=899754846+1=899754847.
#include <bits/stdc++.h>
using namespace std;
namespace fastIO{
	#define BUF_SIZE 100000
	#define OUT_SIZE 100000
	//fread->read
	bool IOerror=0;
	//inline char nc(){char ch=getchar();if(ch==-1)IOerror=1;return ch;} 
	inline char nc(){
		static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
		if(p1==pend){
			p1=buf;pend=buf+fread(buf,1,BUF_SIZE,stdin);
			if(pend==p1){IOerror=1;return -1;}
		}
		return *p1++;
	}
	inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}
	template<class T> inline bool read(T &x){
		bool sign=0;char ch=nc();x=0;
		for(;blank(ch);ch=nc());
		if(IOerror)return false;
		if(ch=='-')sign=1,ch=nc();
		for(;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
		if(sign)x=-x;
		return true;
	}
	inline bool read(double &x){
		bool sign=0;char ch=nc();x=0;
		for(;blank(ch);ch=nc());
		if(IOerror)return false;
		if(ch=='-')sign=1,ch=nc();
		for(;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';
		if(ch=='.'){
			double tmp=1; ch=nc();
			for(;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');
		}
		if(sign)x=-x;
		return true;
	}
	inline bool read(char *s){
		char ch=nc();
		for(;blank(ch);ch=nc());
		if(IOerror)return false;
		for(;!blank(ch)&&!IOerror;ch=nc())*s++=ch;
		*s=0;
		return true;
	}
	inline bool read(char &c){
		for(c=nc();blank(c);c=nc());
		if(IOerror){c=-1;return false;}
		return true; 
	}
	template<class T,class... U>bool read(T& h,U&... t){return read(h)&&read(t...);}
	#undef OUT_SIZE
	#undef BUF_SIZE
};
using namespace fastIO;
/************* debug begin *************/
string to_string(string s){return '"'+s+'"';}
string to_string(const char* s){return to_string((string)s);}
string to_string(const bool& b){return(b?"true":"false");}
template<class T>string to_string(T x){ostringstream sout;sout<<x;return sout.str();}
template<class A,class B>string to_string(pair<A,B> p){return "("+to_string(p.first)+", "+to_string(p.second)+")";}
template<class A>string to_string(const vector<A> v){
	int f=1;string res="{";for(const auto x:v){if(!f)res+= ", ";f=0;res+=to_string(x);}res+="}";
	return res;
}
void debug_out(){puts("");}
template<class T,class... U>void debug_out(const T& h,const U&... t){cout<<" "<<to_string(h);debug_out(t...);}
#ifdef tokitsukaze 
#define debug(...) cout<<"["<<#__VA_ARGS__<<"]:",debug_out(__VA_ARGS__);
#else
#define debug(...) 233;
#endif
/*************  debug end  *************/
#define mem(a,b) memset((a),(b),sizeof(a))
#define MP make_pair
#define pb push_back
#define fi first
#define se second
#define sz(x) (int)x.size()
#define all(x) x.begin(),x.end()
#define sqr(x) (x)*(x)
using namespace __gnu_cxx;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<int,ll> PIL;
typedef pair<ll,int> PLI;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<PII > VPII;
/************* define end  *************/
void read(int *x,int l,int r){for(int i=l;i<=r;i++) read(x[i]);}
void read(ll *x,int l,int r){for(int i=l;i<=r;i++) read(x[i]);}
void read(double *x,int l,int r){for(int i=l;i<=r;i++) read(x[i]);}
void println(VI x){for(int i=0;i<sz(x);i++) printf("%d%c",x[i]," \n"[i==sz(x)-1]);}
void println(VL x){for(int i=0;i<sz(x);i++) printf("%lld%c",x[i]," \n"[i==sz(x)-1]);}
void println(int *x,int l,int r){for(int i=l;i<=r;i++) printf("%d%c",x[i]," \n"[i==r]);}
void println(ll *x,int l,int r){for(int i=l;i<=r;i++) printf("%lld%c",x[i]," \n"[i==r]);}
/*************** IO end  ***************/
void go();
int main(){
	#ifdef tokitsukaze
		freopen("TEST.txt","r",stdin);
	#endif
	go();return 0;
}
const int INF=0x3f3f3f3f;
const ll LLINF=0x3f3f3f3f3f3f3f3fLL;
const double PI=acos(-1.0);
const double eps=1e-6;
const int MAX=2e5+10;
const ll mod=998244353;
/*********************************  head  *********************************/
struct Discretization
{
	#define type ll
	vector<type> a;
	void init(){a.clear();}
	void add(type x){a.pb(x);}
	void work(){sort(all(a));a.resize(unique(all(a))-a.begin());}
	int get_pos(type x){return lower_bound(all(a),x)-a.begin()+1;}
	type get_val(int pos){return a[pos-1];}
	int size(){return a.size();}
	#undef type
}d;
ll x[MAX],y[MAX],bit[MAX];
int s[MAX];
void go()
{
	int n,i,pos;
	ll ans;
	while(read(n))
	{
		d.init();
		for(i=1;i<=n;i++)
		{
			read(x[i],y[i],s[i]);
			d.add(x[i]);
		}
		d.work();
		mem(bit,0);
		ans=x[n]+1;
		for(i=1;i<=n;i++)
		{
			pos=d.get_pos(y[i]);
			bit[i]=(bit[i-1]-bit[pos-1]+mod)%mod;
			bit[i]=(bit[i]+x[i]-y[i])%mod;
			bit[i]=(bit[i-1]+bit[i])%mod;
			if(s[i]) ans=(ans+bit[i]-bit[i-1]+mod)%mod;
		}
		printf("%lld\n",ans);
	}
}
/*
4
3 1 0
6 4 0
7 2 0
10 5 1

*/