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;
int q,_max;
long long n,m,res;
const int P=1e9+7;
inline int mod(long long x){
	return x>=P?x%P:x;
}
void dfs(long long a,long long b,int val,int ty){
	if(a>n||b>m)	return ;
	if(a>m||a+b>n){
		if(val<_max)	return ;
		if(val>_max)	_max=val,res=0;
		if(a<=m)	res=mod(res+(m-a)/b+1);
		_max=max(_max,val),res=mod(res+(n-a)/b+1);
		return ;
	}
	if(!ty)	dfs(a+b,b,val,1);
	dfs(a+b,a,val+1,ty);
}
void solve(){
	if(n==1&&m==1){
		_max=1,res=1;
		return ;
	}
	dfs(2,1,1,0);
	if(_max==1)	res=(n%P*m%P)%P;
}
int main(){
	scanf("%d",&q);
	while(q--){
		scanf("%lld%lld",&n,&m);
		if(n<m)	swap(n,m);
		_max=res=0,solve();
		printf("%d %lld\n",_max,res);
	}
	return 0;
}