Problem: problem

AOR Co., Ltd. is a $ N $ story building. There is no basement floor.
AOR Ika-chan is a squid, so she can go down the stairs, but not up.

I decided to install $ M $ elevators in the building because it would be inconvenient if I couldn't climb upstairs.
It takes time to install the elevator, and the $ i $ th elevator will be installed the night after $ D_i $ and will be mobile on all floors above the $ A_i $ floor and below the $ B_i $ floor.

You were asked $ Q $ questions by AOR Ika-chan. The $ i $ th question is, "Can I move from the $ S_i $ floor to the $ T_i $ floor at noon $ E_i $ days later?"
The only means of transportation are stairs and elevators. In addition, the time required for movement shall be negligible.



input

$ N \ M \ Q $
$ D_1 \ A_1 \ B_1 $
$ \ vdots $
$ D_M \ A_M \ B_M $
$ E_1 \ S_1 \ T_1 $
$ \ vdots $
$ E_Q \ S_Q \ T_Q $

output

Print Yes or No on one line for each question. However, answer in the order in which they are asked. Also, output a line break at the end.

Example

Input

5 1 2
3 1 5
3 1 5
4 1 5


Output

No
Yes

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using Int = long long;

template <typename T,typename E>
struct SegmentTree{
  using F = function<T(T,T)>;
  using G = function<T(T,E)>;
  using H = function<E(E,E)>;
  using P = function<E(E,size_t)>;
  Int n;
  F f;
  G g;
  H h;
  T ti;
  E ei;
  P p;
  vector<T> dat;
  vector<E> laz;
  SegmentTree(Int n_,F f,G g,H h,T ti,E ei,
	      P p=[](E a,size_t b){b++;return a;}):
    f(f),g(g),h(h),ti(ti),ei(ei),p(p){
    init(n_);
  }
  void init(Int n_){
    n=1;
    while(n<n_) n*=2;
    dat.assign(2*n-1,ti);
    laz.assign(2*n-1,ei);
  }
  void build(Int n_, vector<T> v){
    for(Int i=0;i<n_;i++) dat[i+n-1]=v[i];
    for(Int i=n-2;i>=0;i--)
      dat[i]=f(dat[i*2+1],dat[i*2+2]);
  }
  inline void eval(Int len,Int k){
    if(laz[k]==ei) return;
    if(k*2+1<n*2-1){
      laz[k*2+1]=h(laz[k*2+1],laz[k]);
      laz[k*2+2]=h(laz[k*2+2],laz[k]);
    }
    dat[k]=g(dat[k],p(laz[k],len));
    laz[k]=ei;
  }
  T update(Int a,Int b,E x,Int k,Int l,Int r){
    eval(r-l,k);
    if(r<=a||b<=l) return dat[k];
    if(a<=l&&r<=b){
      laz[k]=h(laz[k],x);
      return g(dat[k],p(laz[k],r-l));
    }
    return dat[k]=f(update(a,b,x,k*2+1,l,(l+r)/2),
		    update(a,b,x,k*2+2,(l+r)/2,r));
  }
  T update(Int a,Int b,E x){
    return update(a,b,x,0,0,n);
  }
  T query(Int a,Int b,Int k,Int l,Int r){
    eval(r-l,k);
    if(r<=a||b<=l) return ti;
    if(a<=l&&r<=b) return dat[k];
    T vl=query(a,b,k*2+1,l,(l+r)/2);
    T vr=query(a,b,k*2+2,(l+r)/2,r);
    return f(vl,vr);
  }
  T query(Int a,Int b){
    return query(a,b,0,0,n);
  }
  void update(Int k,T x){
    query(k,k+1);//evaluate
    k+=n-1;
    dat[k]=x;
    while(k){
      k=(k-1)/2;
      dat[k]=f(dat[k*2+1],dat[k*2+2]);
    }
  }
};

//INSERT ABOVE HERE
signed main(){
  Int n,m,q;
  cin>>n>>m>>q;
  using T = tuple<Int, Int, Int, Int>;
  vector<T> v;
  for(Int i=0;i<m;i++){
    Int d,a,b;
    cin>>d>>a>>b;
    v.emplace_back(d,q,a,b);
  }
  for(Int i=0;i<q;i++){
    Int d,a,b;
    cin>>d>>a>>b;
    v.emplace_back(d,i,a,b);
  }
  sort(v.begin(),v.end());
  SegmentTree<Int, Int>::F f=[](Int a,Int b){return max(a,b);};
  SegmentTree<Int, Int> seg(n+1,f,f,f,-1,-1);
  for(Int i=0;i<n+1;i++) seg.update(i,i+1,i);
  
  vector<Int> ans(q);
  for(Int i=0;i<(Int)v.size();i++){
    Int d,x,a,b;
    tie(d,x,a,b)=v[i];
    if(x==q){
      Int k=seg.query(a,b+1);
      seg.update(a,b+1,k);
    }else{
      Int k=seg.query(a,a+1),p=a;
      while(p!=k){
	p=k;
	k=seg.query(a,k+1);
      }
      seg.update(a,k+1,k);
      ans[x]=k>=b;
    }
  }
  for(Int i=0;i<q;i++) cout<<(ans[i]?"Yes":"No")<<endl;
  return 0;
}