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<vector>
#define MD 86 
#define ll long long
#define mod 1000000007
using namespace std;
inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
ll f[MD+5];int ans;
vector<pair<ll,ll> >v[MD+5];
int main()
{
	f[0]=f[1]=1;
	for(int i=2;i<=MD+2;++i) f[i]=f[i-1]+f[i-2];
	v[1].push_back(make_pair(1,2));
	v[1].push_back(make_pair(1,3));
	for(int i=2;i<=MD;++i)
	{
		for(int j=0;j<v[i-1].size();++j)
		{
			ll x=v[i-1][j].first,y=v[i-1][j].second;
			for(x+=y;x<=f[i+2];x+=y) v[i].push_back(make_pair(y,x));
		}
	}
	for(int Q=read();Q--;)
	{
		ll x=read(),y=read();if(x>y) swap(x,y);
		for(ans=1;f[ans+1]<=x&&f[ans+2]<=y;++ans);
		printf("%d ",ans);
		if(ans==1) printf("%d\n",1LL*x*y%mod);
		else 
		{
			int sum=0;
			for(int j=0;j<v[ans-1].size();++j)
			{
				ll X=v[ans-1][j].first,Y=v[ans-1][j].second;
				if(Y<=x&&X+Y<=y) sum=(sum+(y-X)/Y)%mod;
				if(Y<=y&&X+Y<=x) sum=(sum+(x-X)/Y)%mod;
			}
			printf("%d\n",sum);
		}
	}
	return 0;
}