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>
#define LL long long
using namespace std;
int q;
LL n,m;
const int N=105,mod=1e9+7;
vector<pair<LL,LL> >t[N];
LL f[N];
inline LL read()
{
	LL res = 0; char ch = getchar(); bool XX = false;
	for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
	for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
	return XX ? -res : res;
}
void YYCH()
{
	t[1].push_back(make_pair(1,2));
	t[1].push_back(make_pair(1,3));
	t[1].push_back(make_pair(1,4));
	f[0]=f[1]=1;
	for(int i=2;i<=100;++i)f[i]=f[i-1]+f[i-2];
	for(int i=1;i<=100;++i)
		for(int j=0,l=t[i].size()-1;j<=l;++j)
		{
			LL x=t[i][j].second,y=x+t[i][j].first;
			while(y<=f[i+3]+f[i])t[i+1].push_back(make_pair(x,y)),y+=x;
		}
}
signed main()
{
	YYCH();
	cin>>q;
	while(q--)
	{
		n=read();m=read();
		if(n>m)swap(n,m);
		int p=1;LL ans=0;
		while(f[p+1]<=n&&f[p+2]<=m)++p;
		printf("%d ",p);
		if(p==1){printf("%lld\n",n%mod*m%mod);continue;}
		for(int j=0,l=t[p-1].size()-1;j<=l;++j)
		{
			LL x=t[p-1][j].first,y=t[p-1][j].second;
			if(y<=n)(ans+=(m-x)/y)%=mod;
			if(y<=m)(ans+=(n-x)/y)%=mod;
		}
		printf("%lld\n",ans);
	}
	return 0;
}