題目大意：有$n$個位置，$m$個操作。操作有兩種：

1. $1\;l\;r\;x：$在區間$[l,r]$每個位置加上一個數$x$
2. $2\;l\;r\;k：$詢問$[l,r]$中第$k$大的數是多少。

題解：樹套樹，權值線段樹套位置線段樹，要標記永久化，不然會$TLE$

卡點：沒有標記永久化，$TLE$，然後處理$tag$部分寫錯

C++ Code：

#include <cstdio>
#include <algorithm>
#include <cctype>
namespace __IO {
	namespace R {
		int x, ch, f;
		inline int readsign() {
			f = 1;
			while (isspace(ch = getchar()));
			if (ch == '-') f = -1;
			for (x = ch & 15; isdigit(ch = getchar()); ) x = x * 10 + (ch & 15);
			return x * f;
		}
		inline int read() {
			while (isspace(ch = getchar()));
			for (x = ch & 15; isdigit(ch = getchar()); ) x = x * 10 + (ch & 15);
			return x;
		}

		long long X;
		inline long long readll() {
			while (isspace(ch = getchar()));
			for (X = ch & 15; isdigit(ch = getchar()); ) X = X * 10 + (ch & 15);
			return X;
		}
	}
}
using __IO::R::read;
using __IO::R::readsign;
using __IO::R::readll;

#define maxn 50010

int n, m;

namespace SgT2 {
#define N ((maxn << 3) * 50)
	int lc[N], rc[N], tg[N], idx;
	int L, R;
	long long V[N];

	void __insert(int &rt, const int l, const int r) {
		if (!rt) rt = ++idx;
		V[rt] += std::min(R, r) - std::max(L, l) + 1;
		if (L <= l && R >= r) {
			tg[rt]++;
			return ;
		}
		int mid = l + r >> 1;
		if (L <= mid) __insert(lc[rt], l, mid);
		if (R > mid) __insert(rc[rt], mid + 1, r);
	}
	void insert(int &rt, int __L, int __R) {
		L = __L, R = __R;
		__insert(rt, 1, n);
	}

	long long __query(const int rt, const int l, const int r) {
		if (!rt || (L <= l && R >= r)) return V[rt];
		int mid = l + r >> 1;
		long long res = static_cast<long long> (std::min(R, r) - std::max(L, l) + 1) * tg[rt];
		if (L <= mid) res += __query(lc[rt], l, mid);
		if (R > mid) res += __query(rc[rt], mid + 1, r);
		return res;
	}
	long long query(int rt, int __L, int __R) {
		L = __L, R = __R;
		return __query(rt, 1, n);
	}
#undef N
}

namespace SgT {
#define N (maxn << 3)
	const int maxl = -50000, maxr = 50000;
	int root[N];
	int L, R, num;
	void __insert(const int rt, const int l, const int r) {
		SgT2::insert(root[rt], L, R);
		if (l == r) return ;
		int mid = l + r >> 1;
		if (num <= mid) __insert(rt << 1, l, mid);
		else __insert(rt << 1 | 1, mid + 1, r);
	}
	void insert(int __L, int __R, int __num) {
		L = __L, R = __R, num = __num;
		__insert(1, maxl, maxr);
	}

	long long pos;
	int __query(const int rt, const int l, const int r) {
		if (l == r) return l;
		int mid = l + r >> 1;
		long long tmp = SgT2::query(root[rt << 1 | 1], L, R);
		if (pos <= tmp) return __query(rt << 1 | 1, mid + 1, r);
		else {
			pos -= tmp;
			return __query(rt << 1, l, mid);
		}
	}
	int query(int __L, int __R, long long __pos) {
		L = __L, R = __R, pos = __pos;
		return __query(1, maxl, maxr);
	}
#undef N
}
using SgT::insert;
using SgT::query;

int main() {
	n = read(), m = read();
	while (m --> 0) {
		int op = read(), l = read(), r = read();
		if (op == 1) {
			int c = readsign();
			insert(l, r, c);
		} else {
			long long c = readll();
			printf("%d\n", query(l, r, c));
		}
	}
	return 0;
}