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>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a < b) ? (a = b, 1) : 0;}
template<class T>bool chmin(T &a, const T &b) { return (b < a) ? (a = b, 1) : 0;}

using ll = long long;
using R = long double;
const R EPS = 1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r) {return (r > EPS) - (r < -EPS);}
inline R sq(R x) {return sqrt(max(x, 0.0L));}

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// Problem Specific Parameter:

// Description: セグメント木　点更新 区間クエリ
// TimeComplexity: 初期化$\mathcal{O}(n \log n)$ 更新とクエリ$\mathcal{O}(\log n)$
// Verifyed: AOJ DSL_2_A

template <int depth> struct Segment_tree {
	const static int h = depth;
	const static int n = 1 << h;

	using T = long long;
	T data[2 * n];
	const T out = 0LL;

	inline T vmerge(T l, T r) {return max(l, r);}

	void init() {
		fill_n(data, 2 * n, out);
	}

	void update(int p, T x) { // set value at position p
		for (data[p += n] = x; p > 1; p >>= 1) {
			data[p >> 1] = vmerge(data[p], data[p ^ 1]);
		}
	}

	T query(int l, int r) {  // sum on interval [l, r)
		T resl = out, resr = out;
		for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
			if (l & 1) resl = vmerge(data[l++], resl);
			if (r & 1) resr = vmerge(resr, data[--r]);
		}
		return vmerge(resl, resr);
	}
};

const int limit = 100010;
const int inf = 1 << 30;
int ary[limit];

multiset<int> st;
vector<int> s[limit], t[limit];

Segment_tree<17> seg;

int main(void) {
	int n, m, q;
	cin >> n >> m >> q;

	st.insert(inf);

	rep(i, m) {
		int d, a, b;
		cin >> d >> a >> b;
		a--, b--;
		s[a].push_back(d);
		t[b].push_back(d);
	}

	rep(i, n - 1) {
		for (auto &elem : t[i]) {
			auto it = st.find(elem);
			st.erase(it);
		}

		for (auto &elem : s[i]) {
			st.insert(elem);
		}

		ary[i] = *begin(st);
		//cerr << i << " " <<  ary[i] << endl;
	}

	seg.init();

	rep(i, n - 1) seg.update(i, ary[i]);

	rep(loop, q) {
		int e, a, b;
		cin >> e >> a >> b;
		a--, b--;

		//cerr << seg.query(a, b) << endl;

		if (a >= b or seg.query(a, b) < e) {
			puts("Yes");
		} else {
			puts("No");
		}
	}

	return 0;
}

// 1 2 3