Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings.

The entire horizon is represented by a number line with N (1 ≤ N ≤ 40,000) buildings. Building i's silhouette has a base that spans locations A_i through B_i along the horizon (1 ≤ A_i < B_i ≤ 1,000,000,000) and has height H_i (1 ≤ H_i ≤ 1,000,000,000). Determine the area, in square units, of the aggregate silhouette formed by all N buildings.

Line 1: The total area, in square units, of the silhouettes formed by all N buildings

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <algorithm>

using namespace std;

int read() {

	int x = 0, f = 1; char c = getchar();

	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }

	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }

	return x * f;

}

#define maxn 40010

#define LL long long

int n, cnt, ca, cd;

struct Rec {

	LL x1, y1, x2, y2;

	Rec() {}

	Rec(LL _1, LL _2, LL _3, LL _4): x1(_1), y1(_2), x2(_3), y2(_4) {}

} rs[maxn];

struct Rec_int { int x1, y1, x2, y2; } rsi[maxn];

LL num[maxn<<1], A[maxn<<1], B[maxn<<1];

struct Line {

	int l, r, x;

	Line() {}

	Line(int _1, int _2, int _3): l(_1), r(_2), x(_3) {}

	bool operator < (const Line& t) const { return x < t.x; }

} ad[maxn], de[maxn];

LL sumv[maxn<<3];

int addv[maxn<<3];

void build(int L, int R, int o) {

	if(L == R) {

		sumv[o] = 0;

		addv[o] = 0;

		return ;

	}

	int M = L + R >> 1, lc = o << 1, rc = lc | 1;

	build(L, M, lc); build(M+1, R, rc);

	sumv[o] = 0; addv[o] = 0;

	return ;

}

void update(int L, int R, int o, int ql, int qr, int v) {

	int M = L + R >> 1, lc = o << 1, rc = lc | 1;

	if(ql <= L && R <= qr) {

		addv[o] += v;

		if(addv[o]) sumv[o] = A[R] - A[L-1];

		else if(L == R) sumv[o] = 0;

		else sumv[o] = sumv[lc] + sumv[rc];

		return ;

	}

	if(ql <= M) update(L, M, lc, ql, qr, v);

	if(qr > M) update(M+1, R, rc, ql, qr, v);

	sumv[o] = addv[o] ? A[R] - A[L-1] : sumv[lc] + sumv[rc];

	return ;

}

int main() {

	int kase = 0;

	n = read();

	cnt = 0;

	for(int i = 1; i <= n; i++) {

		int l = read(), r = read(), h = read();

		rs[i] = Rec(l, 0, r, h);

		num[++cnt] = l; num[++cnt] = r;

	}

	sort(num + 1, num + cnt + 1);

	cnt = unique(num + 1, num + cnt + 1) - num - 1;

	int tcnt = cnt;

	for(int i = 1; i < cnt; i++) A[i] = num[i+1] - num[1];

	for(int i = 1; i <= n; i++) {

		rsi[i].x1 = lower_bound(num + 1, num + cnt + 1, rs[i].x1) - num;

		rsi[i].x2 = lower_bound(num + 1, num + cnt + 1, rs[i].x2) - num - 1;

	}

	cnt = 0;

	for(int i = 1; i <= n; i++)

		num[++cnt] = rs[i].y1, num[++cnt] = rs[i].y2;

	sort(num + 1, num + cnt + 1);

	cnt = unique(num + 1, num + cnt + 1) - num - 1;

	for(int i = 1; i < cnt; i++) B[i] = num[i+1] - num[i];

	ca = cd = 0;

	for(int i = 1; i <= n; i++) {

		rsi[i].y1 = lower_bound(num + 1, num + cnt + 1, rs[i].y1) - num;

		rsi[i].y2 = lower_bound(num + 1, num + cnt + 1, rs[i].y2) - num;

		ad[++ca] = Line(rsi[i].x1, rsi[i].x2, rsi[i].y1);

		de[++cd] = Line(rsi[i].x1, rsi[i].x2, rsi[i].y2);

	}

	sort(ad + 1, ad + ca + 1);

	sort(de + 1, de + cd + 1);

	LL ans = 0;

	int ka = 1, kd = 1;

	build(1, tcnt, 1);

	for(int i = 1; i <= cnt; i++) {

		while(ka <= ca && ad[ka].x == i)

			update(1, tcnt, 1, ad[ka].l, ad[ka].r, 1), ka++;

		while(kd <= cd && de[kd].x == i)

			update(1, tcnt, 1, de[kd].l, de[kd].r, -1), kd++;

		if(i < cnt) ans += sumv[1] * B[i];

	}

	printf("%lld\n", ans);

	return 0;

}