/**
 * We will solve the problem for each bit separately.
 * For a fixed bit, the reccurence G[r, c] = G[r - 1, c] xor G[r, c - 1] is
 * the same as used in Sierpinski triangle (Pascal's triangle modulo 2), the
 * only difference is the base condition. The values in first row/column are
 * very regular in our case, first row has only zeroes (or only ones), and
 * first column has 2^k zeroes, 2^k ones, 2^k zeroes, 2^k ones, and so on.
 * We can notice that the column number 2^k of the original Sierpinski triangle
 * has exactly the same pattern. Another thing about the original Sierpinski
 * triangle is that the row number (2^k - 1) has the following structure:
 *         1 0 0 0 0 0 ... 0 0 0
 *           <--- (2^k - 1) --->
 * So, our Xor Grid (for k-th bit) is just a part of the Sierpinski triangle.
 *
 * Then we will use the fact that Sierpinski triangle is a fractal and to find
 * the xor sum of a rectangle, we just want to find the parity of number of ones.
 **/

#include<bits/stdc++.h>

using namespace std;

// Calculates the i-th bit of G[-inf..r, -inf..c]
long long f(long long r, long long c, int i) {
	r += (1ll << 62) - (1ll << i) + 1;
	c += (1ll << i);
	
	if (r & c) return (1ll << i);
	
	return 0;
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		long long r1, r2, c1, c2;
		scanf("%lld %lld %lld %lld", &r1, &r2, &c1, &c2);
		r1--;
		c1--;
		long long ans = 0;
		for (int i = 0; i < 60; i++) {
			ans ^= f(r2, c2, i);
			ans ^= f(r1, c2, i);
			ans ^= f(r2, c1, i);
			ans ^= f(r1, c1, i);
		}
		
		printf("%lld\n", ans);
	}
	
	return 0;
}