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 <cstdio>
#include <algorithm>

using namespace std;

const int p=1000000007;

long long f[200],f_[200];
long long m,n;
int h,i,s,t;

inline void dfs(int x,int y,long long u,long long v,long long su,long long sv)
{
	if (x==y-1)
	{
		s=(s+(m-u)/v)%p;
		s=(s+(n-u)/v)%p;
	}
	else
	{
		long long tu,tv,tw;
		for (int k=1;;k++)
		{
			tu=v,tv=k*v+u;
			tu=su+(k-1)*v*f_[y-x-2];
			tv=sv+(k-1)*v*f_[y-x-1];
			if ((tu>n) || (tv>m))
				break;
			dfs(x+1,y,v,k*v+u,tu,tv);
		}
	}
	return;
}

int main()
{
	f[0]=0,f[1]=1,f[2]=2;
	for (i=3;i<=100;i++)
		f[i]=f[i-1]+f[i-2];
	f_[0]=1,f_[1]=1;
	for (i=2;i<=100;i++)
		f_[i]=f_[i-1]+f_[i-2];
	scanf("%d",&t);
	for (h=1;h<=t;h++)
	{
		scanf("%lld%lld",&n,&m);
		if (n>m)
			swap(n,m);
		if ((n==1) && (m==1))
		{
			printf("%d %d\n",1,1);
			continue;
		}
		for (i=0;(f[i]<=n) && (f[i+1]<=m);i++);
		printf("%d ",i-1);
		if (i-1==1)
			printf("%lld\n",(m%p)*(n%p)%p);
		else
		{
			s=0;
			dfs(0,i-1,1,1,f[i-1],f[i]);
			printf("%d\n",s);
		}
	}
	return 0;
}