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:
/*deco loves Chino*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
int fib[1100];
vector<pair<int,int> > ans[1100];
const int mod=1e9+7;
signed main()
{
	fib[0]=fib[1]=1;
	for(int i=2;i<=90;i++)
	{
		fib[i]=fib[i-1]+fib[i-2];
	}
	ans[1].push_back(make_pair(1,2)),ans[1].push_back(make_pair(1,3)),ans[1].push_back(make_pair(1,4));
	for(int i=1;i<=87;i++)
	{
		for(int j=0;j<ans[i].size();j++)
		{
			pair<int,int> qwq=ans[i][j];
			int dx=qwq.second,dy=qwq.first+dx;
			while(dy<=fib[i]+fib[i+3])
			{
				ans[i+1].push_back(make_pair(dx,dy));
				dy+=dx;	
			} 
		}
	}
	int t;
	cin>>t;
	while(t--)
	{
		int n,m;
		scanf("%lld%lld",&n,&m);
		if(n>m)
		{
			swap(n,m);
		}
		int now=1;
		while(fib[now+1]<=n&&fib[now+2]<=m)
		{
			++now;
		}
		cout<<now<<" ";
		if(now==1)
		{
			cout<<n*m%mod<<"\n";
			continue;
		}
		int anss=0;
		for(int i=0;i<ans[now-1].size();i++)
		{
			pair<int,int> qwq=ans[now-1][i];
			if(qwq.second<=m)
			{
				anss+=((n-qwq.first)/qwq.second)%mod;
			}
			if(qwq.second<=n)
			{
				anss+=((m-qwq.first)/qwq.second)%mod;
			}
			anss%=mod;
		}
		cout<<anss<<"\n";
	} 
}