#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;

struct Node {
	int l, r, d;
}node[N];
int a[N];
int n, m, k, t;
map<int, int> mp;

bool check(int x, int y) {
	int T = 0;
	mp.clear();
	for (int i = 1; i <= k; i++) {
		if (node[i].d > y) { // 如果该段危险值大于下限，加上这段
			mp[node[i].l]++;
			mp[node[i].r+1]--;
		}
	}
	int pre = 0; // 前缀和
	for (int i = 1; i <= n+1; i++) {
		pre += mp[i];
		if (pre >= 1) { // 如果当前位置大于等于1，说明要先摧毁陷阱，故在这个点上花费的时间为3
			T += 3;
		} else {
			T++;
		}
	}
	return T <= t;
}


int main() {

	cin >> m >> n >> k >> t;
	for (int i = 1; i <= m; i++) cin >> a[i];
	for (int i = 1; i <= k; i++) {
		cin >> node[i].l >> node[i].r >> node[i].d;
	}
	sort(a+1, a+1+m, greater<int>());
	int l = 0, r = m, mid;
	while (l < r) {
		mid = (l+r+1) >> 1; // 这里要注意向上取整，即相邻的数优先判断大的是否满足条件
		if (check(mid, a[mid])) {
			l = mid;
		} else {
			r = mid - 1;
		}

	}
	cout << l << endl;

	return 0;
}