#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define RFOR(i, a, b) for(int i = (a) - 1; i >= (b); i--)
#define SZ(a) int(a.size())
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
#define F first
#define S second

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef double db;

const int mod = 1e9 + 7;

int add(int a, int b)
{
	return a + b < mod ? a + b : a + b - mod;
}
int mult(int a, int b)
{
	return (LL)a * b % mod;
}

const int N = 30;
int dp[N][2][2];
int x, y;

int f(int len, int ex, int ey)
{
	if (len == N)
		return 1;
	if (dp[len][ex][ey] != -1)
		return dp[len][ex][ey];
	dp[len][ex][ey] = 0;
	int btx = ((x >> (N - len - 1)) & 1);
	int bty = ((y >> (N - len - 1)) & 1);
	FOR (bx, 0, 2)
	{
		if (ex && bx > btx)
			continue;
		int nex = ex && bx == btx;
		FOR (by, 0, 2)
		{
			if (ey && by > bty)
				continue;
			if (ey == ex && ey == 1)
				continue;
			int ney = ey && by == bty;
			add(dp[len][ex][ey], f(len + 1, nex, ney));
		}
	}
	return dp[len][ex][ey];
}

void solve()
{
	FOR (i, 0, N) FOR (j, 0, 2) FOR (k, 0, 2) dp[i][j][k] = -1;
	
	cin >> x >> y;
	
	int eqx = 1;
	int eqy = 1;
	
	int ans = 0;
	
	FOR (i, 0, N)
	{
		int btx = (x >> (N - 1 - i)) & 1;
		int bty = (y >> (N - 1 - i)) & 1;
		FOR (bx, 0, 2)
		{
			if (eqx && bx > btx)
				continue;
			FOR (by, 0, 2)
			{
				if (eqy && by > bty)
					continue;
				if (bx == by)
					continue;
				ans = add(ans, mult(N - i, f(i + 1, eqx && bx == btx, eqy && by == bty)));
			}
		}
		eqx &= btx == 0;
		eqy &= bty == 0;
	}
	cout << ans << '\n';
}

int main()
{
	ios::sync_with_stdio(0); 
	cin.tie(0);
	cout << fixed << setprecision(15);
	
	int t;
	cin >> t;
	while (t--)
	{
		solve();
	}
	
	return 0;
}