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>
using namespace std;

typedef long long LL;

const int Mod=1e9+7;

int N,M,Ans,Cnt;
LL F[90];

vector<pair<LL,LL>> V[105];

int main(){
	int i;
	LL x,y,p,q;
	pair<int,int> t;
	F[0]=F[1]=1;
	for(i=2;;i++){
		F[i]=F[i-1]+F[i-2];
		if(F[i]>1e18)
			break;
	}
	N=i;
	for(i=2;i<5;i++)
		V[1].push_back(make_pair(1,i));
	for(i=1;i+3<=N;i++)
		for(auto t:V[i]){
			x=t.second,y=t.first+x;
			for(;y<=F[i+3]+F[i];y+=x)
				V[i+1].push_back(make_pair(x,y));
		}
	scanf("%d",&M);
	while(M--){
		scanf("%lld%lld",&p,&q);
		if(p>q)
			swap(p,q);
		for(Ans=1;F[Ans+1]<=p&&F[Ans+2]<=q;Ans++);
		printf("%d ",Ans);
		if(Ans==1){
			printf("%d\n",1llu*(p%Mod)*(q%Mod)%Mod);
			continue;
		}
		Cnt=0;
		for(auto t:V[Ans-1]){
			x=t.first,y=t.second;
			if(y<=p)
				Cnt=(Cnt+(q-x)/y)%Mod;
			if(y<=q)
				Cnt=(Cnt+(p-x)/y)%Mod;
		}
		printf("%d\n",Cnt);
	}
	return 0;
}