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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define int long long

const int MAX=88;
const int mod=1e9+7;

inline int read()
{
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9')
	{
		if(c=='-')
			f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}

int A,B;
int re1[MAX+5],re2[MAX+5];
int len;
int k1,k2;

int Work(int x,int y,int tot)
{
	if(x>A||y>B)
		return tot-1;
	if(x!=y)
	{
		re1[++len]=x,re2[len]=y;
		k1=x,k2=y;
	}
	return Work(x+y,x,tot+1);
}

int f[MAX+5];
int sum;

void Find1(int x,int val1,int val2)
{
	int sum1=0,sum2;
	if(!val2)
		sum2=0x3f3f3f3f3f3f3f3f;
	else
		sum2=(B-k2)/re2[x]/val2;
	sum1=((A-k1)/re2[x])/val1;
	for(int i=1;i<=min(sum1,sum2);i++)
	{
		int t1=k1+i*val1*re2[x];
		int t2=k2+i*val2*re2[x];
		sum=(sum+(A-t1)/t2+1)%mod;
		if(B>=t1)
			sum=(sum+1)%mod;
	}
}

signed main()
{
	f[1]=1;
	f[2]=1;
	for(int i=3;i<=MAX;i++)
		f[i]=f[i-1]+f[i-2];
	int T=read();
	while(T--)
	{
		A=read(); B=read();
		len=0;
		if(A<B)
			swap(A,B);
		if(A==1&&B==1)
		{
			cout<<"1"<<" "<<"1"<<endl;
			continue;
		}
		if(A==2&&B==1)
		{
			cout<<"1"<<" "<<"2"<<endl;
			continue;
		}
		if(B==1)
		{
			cout<<"1"<<" "<<A%mod<<endl;
			continue;
		}
		int ans1=Work(1,0,0)-1;
		printf("%lld ",ans1);
		if(A==2&&B==2)
		{
			cout<<"4"<<endl;
			continue;
		}
		if(k1<=B)
			sum=2;
		else
			sum=1;
		sum=(sum+(A-k1)/k2)%mod;
		for(int i=2;i<len;i++)
		{
			int val1=f[len-i+1];
			int val2=f[len-i];
			Find1(i,val1,val2);
		}
		printf("%lld\n",sum);
	}
	return 0;
}