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<cstdio>
#include<set>
#define inf 1e18
#define mk make_pair
#define fi first
#define se second
#define pii pair<long long,long long>
#define p 1000000007
using namespace std;
int T,sz;
long long ans;
long long f[1010],b[1010];
set<pii> s[1010];
set<pii>::iterator it;
int main()
{
	scanf("%d",&T);
	f[0]=f[1]=1;
	for(sz=2;;sz++)
	{
		f[sz]=f[sz-1]+f[sz-2];
		if(f[sz]>=inf) break;
	}
	for(int i=1;i<sz;i++)
		if(f[i]*2>f[i+1])
		s[i].insert(mk(f[i],f[i+1]));
	s[1].insert(mk(1,1));
	for(int i=1;i<=sz-2;i++)
	{
		b[i]=f[i],b[i+1]=f[i+2];
		if(b[i]*2LL>b[i+1])
		s[i].insert(mk(b[i],b[i+1]));
		for(int j=i+1;;j++)
		{
			b[j+1]=b[j]+b[j-1];
			if(b[j+1]>=inf) break;
			if(b[j]*2LL>b[j+1])
			s[j].insert(mk(b[j],b[j+1]));
		}
	}
	while(T--)
	{
		long long X,Y;
		scanf("%lld%lld",&X,&Y);
		if(X>Y) swap(X,Y);
		int ma=1;
		for(;;ma++)
		{
			pii x=*s[ma].begin();
			if(x.fi>X||x.se>Y) break;
		}
		ma--;
		if(ma==1) ans=X%p*Y%p;
		else
		for(it=s[ma].begin();it!=s[ma].end();it++)
		{
			pii x=*it;
			if(x.fi<=X)
				(ans+=max(0LL,Y-x.se+x.fi)/x.fi+max(0LL,X-x.se+x.fi)/x.fi)%=p;
		}
		printf("%d %lld\n",ma,ans);
		ans=0;
	}			
	return 0;
}