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;
typedef long long LL;

const int Mod=1e9+7;

int m;
LL fib[100];
vector<pair<LL,LL> > good[100];

void Getgoodnumber(int x)
{
	for (auto d:good[x-1])
	{
		LL add=d.first; swap(d.first,d.second);
		for (d.first+=add;d.first<=fib[x+2];d.first+=add)
			good[x].push_back(d);
	}
	sort(good[x].begin(),good[x].end());
	good[x].erase(unique(good[x].begin(),good[x].end()),good[x].end());
}

void FindAns(LL a,LL b)
{
	int ans=1;
	for (int i=2;i<=90;++i)
		if (fib[i]<=a && fib[i+1]<=b) ans=i;
		else break;
	if (ans==1)
		printf("%d %lld\n",ans,1ll*(a%Mod)*(b%Mod)%Mod);
	else 
	{
		LL ret=0;
		for (auto d:good[ans-1])
		{
			if (d.first<=a) ret=(ret+(b-d.second)/d.first)%Mod;
			if (d.first<=b) ret=(ret+(a-d.second)/d.first)%Mod;
		}
		printf("%d %lld\n",ans,ret);
	}
}

void Solve()
{
	fib[0]=fib[1]=1; for (int i=2;i<=91;++i) fib[i]=fib[i-1]+fib[i-2];
	good[1].push_back(make_pair(2,1)); good[1].push_back(make_pair(3,1));
	for (int i=2;i<=91;++i) Getgoodnumber(i);
	scanf("%d",&m); LL a,b;
	for (int i=1;i<=m;++i)
	{
		scanf("%lld%lld",&a,&b);
		if (a>b) swap(a,b);
		FindAns(a,b);
	}
}

int main()
{
	Solve();
	return 0;
}