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 rep(i,n) for(int i=0;i<n;i++)
#define lint long long
using namespace std;
lint q,x,y,f[100]={0,1,1},m=1000000007;
int main(){
	cin>>q;
	rep(i,97)f[i+3]=f[i+1]+f[i+2];
	rep(i,q){
		cin>>x>>y;
		if(x>y)swap(x,y);
		if(y<=2){
			cout<<1<<" ";
			cout<<(y==1?1:x==1?2:4)<<"\n";
			continue;
		}
		rep(j,100){
			if(x<f[j]||y<f[j+1]){
				cout<<j-2<<" ";
				lint a=x==1?2:1;
				rep(k,j-2){
					lint n=f[j-1]+f[j-3-k]*f[k+2],m=f[j]+f[j-2-k]*f[k+2];
					if(n<=x&&m<=y)a+=(y-m)/n+1;
				}
				if(f[j]<=x){
					a++;
					rep(k,j-2)a+=(x-f[j])/(f[j-2-k]*f[k+2]);
				}
				cout<<a%m<<"\n";
				break;
			}
		}
	}
}