Given a set of $N$ axis-aligned rectangles in the plane, find the area of regions which are covered by at least one rectangle.

Constraints

* $ 1 \leq N \leq 2000 $
* $ −10^9 \leq x1_i < x2_i\leq 10^9 $
* $ −10^9 \leq y1_i < y2_i\leq 10^9 $

Input

The input is given in the following format.

$N$
$x1_1$ $y1_1$ $x2_1$ $y2_1$
$x1_2$ $y1_2$ $x2_2$ $y2_2$
:
$x1_N$ $y1_N$ $x2_N$ $y2_N$


($x1_i, y1_i$) and ($x2_i, y2_i$) are the coordinates of the top-left corner and the bottom-right corner of the $i$-th rectangle respectively.

Output

Print the area of the regions.

Examples

Input

2
0 0 3 4
1 2 4 3


Output

13


Input

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


Output

7


Input

4
0 0 3 1
0 0 1 3
0 2 3 3
2 0 3 3


Output

8
# include "bits/stdc++.h"
using namespace std;
using LL = long long;
using ULL = unsigned long long;
const double PI = acos(-1);
template<class T>constexpr T INF() { return ::std::numeric_limits<T>::max(); }
template<class T>constexpr T HINF() { return INF<T>() / 2; }
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); };
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); };
typedef pair<LL, LL> pii;
const int vy[] = { -1, -1, -1, 0, 1, 1, 1, 0 }, vx[] = { -1, 0, 1, 1, 1, 0, -1, -1 };
const int dx[4] = { -1,0,1,0 }, dy[4] = { 0,-1,0,1 };
const char dir[4] = { 'u','l','d','r' };
int popcnt(unsigned long long n) { int cnt = 0; for (int i = 0; i < 64; i++)if ((n >> i) & 1)cnt++; return cnt; }
int d_sum(LL n) { int ret = 0; while (n > 0) { ret += n % 10; n /= 10; }return ret; }
int d_cnt(LL n) { int ret = 0; while (n > 0) { ret++; n /= 10; }return ret; }
LL gcd(LL a, LL b) { if (b == 0)return a; return gcd(b, a%b); };
LL lcm(LL a, LL b) { LL g = gcd(a, b); return a / g*b; };
# define ALL(qpqpq)           (qpqpq).begin(),(qpqpq).end()
# define UNIQUE(wpwpw)        sort(ALL((wpwpw)));(wpwpw).erase(unique(ALL((wpwpw))),(wpwpw).end())
# define LOWER(epepe)         transform(ALL((epepe)),(epepe).begin(),TL<char>)
# define UPPER(rprpr)         transform(ALL((rprpr)),(rprpr).begin(),TU<char>)
# define FOR(i,tptpt,ypypy)   for(LL i=(tptpt);i<(ypypy);i++)
# define REP(i,upupu)         FOR(i,0,upupu)
# define INIT                 std::ios::sync_with_stdio(false);std::cin.tie(0)
# pragma warning(disable:4996)

int n;
LL x1[2020], x2[2020], y334[2020], y2[2020];
int iscover[4040][4040];
int main() {
	cin >> n;
	vector<LL>x, y;
	REP(i, n) {
		cin >> x1[i] >> y334[i] >> x2[i] >> y2[i];
		x.emplace_back(x1[i]);
		x.emplace_back(x2[i]);
		y.emplace_back(y334[i]);
		y.emplace_back(y2[i]);
	}
	UNIQUE(x);
	UNIQUE(y);
	REP(i, n) {
		int xx1 = lower_bound(ALL(x), x1[i]) - x.begin();
		int xx2 = lower_bound(ALL(x), x2[i]) - x.begin();
		int yy1 = lower_bound(ALL(y), y334[i]) - y.begin();
		int yy2 = lower_bound(ALL(y), y2[i]) - y.begin();
		iscover[yy1][xx1]++;
		iscover[yy2][xx1]--;
		iscover[yy1][xx2]--;
		iscover[yy2][xx2]++;
	}
	REP(i, 4040) {
		REP(j, 4040) {
			if (j)iscover[i][j] += iscover[i][j - 1];
		}
	}
	REP(j, 4040) {
		REP(i, 4040) {
			if (i)iscover[i][j] += iscover[i - 1][j];
		}
	}
	LL ans = 0;
	REP(i, 4040)REP(j, 4040) {
		if (iscover[i][j] > 0) {
			ans += (x[j + 1] - x[j])*(y[i + 1] - y[i]);
		}
	}
	cout << ans << endl;
	//system("pause");
}
