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 <bits/stdc++.h>

using namespace std;

#define M 1005
#define MOD 1000000007
#define lim 1e18
#define ll long long
#define mp make_pair
#define fi first
#define se second

int k = 1, T, mx;
ll x, y, res, fib[M];

vector< pair<ll, ll> > exellent[M];

inline ll read()
{
	ll x = 0, f = 1; char ch = getchar();
	while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
	while (isdigit(ch)) { (x *= 10) += ch - '0'; ch = getchar(); }
	return x * f;
}

int main()
{
	fib[0] = fib[1] = 1;
	while (fib[k] + fib[k-1] <= lim)
	{	k++; fib[k] = fib[k-1] + fib[k-2]; }

	exellent[1].push_back(mp(1, 2));
	for (int i = 2; i <= k; i++)
	{
		for (pair<ll, ll> p : exellent[i-1])
			exellent[i].push_back(mp(p.se, p.fi + p.se));
		exellent[i].push_back(mp(fib[i+1], fib[i+1] + fib[i-1]));
	}

	T = read();
	while (T--)
	{
		x = read(), y = read(); mx = 1, res = 0;
		if (x > y) swap(x, y);
		while (mx+2 <= k && fib[mx+1] <= x && fib[mx+2] <= y) mx++;
		for (pair<ll, ll> p : exellent[mx])
		{
			if (p.fi <= x && p.se <= y)
				(res += (y - p.se) / p.fi + 1) %= MOD;
			if (p.se <= x)
				(res += (x - p.se) / p.fi + 1) %= MOD;
		}
		if (mx == 1) res += x;
		printf("%d %lld\n", mx, res);
	}

	return 0;
}