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;
#define mo 1000000007ll
#define N 110
#define int long long
struct no{
	int x,y;
};
vector<no>v[N];
int f[N],q;
signed main(){
	v[1].push_back((no){1,2});
	v[1].push_back((no){1,3});
	v[1].push_back((no){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<=97;i++)
		for(auto x:v[i]){
			int a=x.y,b=x.x+a;
			while(b<=f[i+3]+f[i])
				v[i+1].push_back((no){a,b}),b+=a;
		}
	cin>>q;
	while(q--){
		int n,m;
		cin>>n>>m;
		if(n>m)swap(n,m);
		int g=1,r=0;
		while(f[g+1]<=n&&f[g+2]<=m)g++;
		cout<<g<<' ';
		if(g==1)cout<<n%mo*m%mo<<'\n';
		else{
			for(auto x:v[g-1]){
			int a=x.x,b=x.y;
			if(b<=n)r=(r+(m-a)/b)%mo;
			if(b<=m)r=(r+(n-a)/b)%mo;
		}
		cout<<r<<'\n';
		}
	}
}