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>
#define int long long
#define mod 1000000007
using namespace std;
int q, x, y, f[107], sum;
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int dfs(int a, int b, int d)
{
	if (a * f[d - 1] + b * (d > 1 ? f[d - 2] : 0) > x || a * f[d] + b * f[d - 1] > y)
		return 0;
	if (d == 1)
	{
		if (x >= a + b)sum = (sum + (x - b) / a % mod) % mod;
		sum = (sum + (y - b) / a % mod) % mod;
		return 1;
	}
	int i = 1 + (b == 0);
	while (1)
	{
		if (!dfs(b + i * a, a, d - 1))return i > 1;
		i++;
	}
}
signed main()
{
	q = read();
	f[0] = f[1] = 1;
	for (int i = 2; i <= 88; i++)
		f[i] = f[i - 1] + f[i - 2];
	for (int lp = 1; lp <= q; lp++)
	{
		x = read();
		y = read();
		if (x > y)swap(x, y);
		if (x == 1) { printf("1 %lld\n", y % mod); continue; }
		int res = 0;
		while (x >= f[res] && y >= f[res + 1])
			res++;
		res--;
		sum = 0;
		dfs(1, 0, res);
		printf("%lld %lld\n", res, sum);
	}
	return 0;
}