Problem: Kenus, the organizer of International Euclidean Olympiad, is seeking a pair of two integers that requires many steps to find its greatest common divisor using the Euclidean algorithm.

You are given Q queries. The i-th query is represented as a pair of two integers X_i and Y_i, and asks you the following: among all pairs of two integers (x,y) such that 1 ≤ x ≤ X_i and 1 ≤ y ≤ Y_i, find the maximum Euclidean step count (defined below), and how many pairs have the maximum step count, modulo 10^9+7.

Process all the queries. Here, the Euclidean step count of a pair of two non-negative integers (a,b) is defined as follows:

* (a,b) and (b,a) have the same Euclidean step count.
* (0,a) has a Euclidean step count of 0.
* If a > 0 and a ≤ b, let p and q be a unique pair of integers such that b=pa+q and 0 ≤ q < a. Then, the Euclidean step count of (a,b) is the Euclidean step count of (q,a) plus 1.

Constraints

* 1 ≤ Q ≤ 3 × 10^5
* 1 ≤ X_i,Y_i ≤ 10^{18}(1 ≤ i ≤ Q)

Input

The input is given from Standard Input in the following format:


Q
X_1 Y_1
:
X_Q Y_Q


Output

For each query, print the maximum Euclidean step count, and the number of the pairs that have the maximum step count, modulo 10^9+7, with a space in between.

Examples

Input

3
4 4
6 10
12 11


Output

2 4
4 1
4 7


Input

10
1 1
2 2
5 1000000000000000000
7 3
1 334334334334334334
23847657 23458792534
111111111 111111111
7 7
4 19
9 10


Output

1 1
1 4
4 600000013
3 1
1 993994017
35 37447
38 2
3 6
3 9
4 2

Solution (in CPP):

Solution:
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

typedef long long s64;

const int L = 86;
const int M = 1000000007;

int nQ;

s64 f[L + 2];

struct object
{
	s64 x, y;
	object() {}
	object(const s64 &_x, const s64 &_y)
		: x(_x), y(_y) {}
};
vector<object> ex[L + 1];

int main()
{
	ios::sync_with_stdio(false);

	f[0] = f[1] = 1;
	for (int i = 2; i <= L + 1; ++i)
		f[i] = f[i - 1] + f[i - 2];

	ex[1].emplace_back(1, 2);
	ex[1].emplace_back(1, 3);
	ex[1].emplace_back(1, 4);
	for (int i = 1; i <= L - 3; ++i)
	{
		s64 lim = f[i] + f[i + 3];

		for (object k : ex[i])
		{
			s64 x = k.x, y = k.y;
			while ((x += y) <= lim)
				ex[i + 1].emplace_back(y, x);
		}
	}

	cin >> nQ;
	while (nQ--)
	{
		s64 n, m;
		cin >> n >> m;

		if (n > m)
			swap(n, m);

		int l = 1;
		while (f[l + 1] <= n && f[l + 2] <= m)
			++l;
		cout << l << ' ';

		if (l == 1)
			cout << (s64)n * m % M << endl;
		else
		{
			int res = 0;
			for (object k : ex[l - 1])
			{
				s64 x = k.x, y = k.y;
				if (y <= n)
					res = (res + (m - x) / y) % M;
				if (y <= m)
					res = (res + (n - x) / y) % M;
			}

			cout << res << endl;
		}
	}

	return 0;
}