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<iostream>
#define m long long
using namespace std;
#define z else 
#define d(s,i,n) for(m i=s;i<n;i++)
m l[90],r[90],e[90][90],f[90][90];
int main(){m n;cin>>n;l[0]=1;r[0]=1;d(1,i,90)l[i]=r[i-1],r[i]=l[i-1]+r[i-1];d(0,i,89){e[i][0]=1,f[i][0]=1;d(1,j,90){if(i+1==j)f[i][j]=f[i][j-1]*2+e[i][j-1];z f[i][j]=f[i][j-1]+e[i][j-1];e[i][j]=f[i][j-1];}}d(0,i,n){m c,d;cin>>c>>d;if(d>c)swap(c,d);m k=0;d(0,j,90)if(c>=r[j]&&d>=l[j])k=j;m g;if(k==0)k=1,g=1;z if(k==1)if(d==1)g=c;z g=4;z{g=1;if(d>=r[k]&&c>=l[k]&&l[k]<r[k])g++;d(0,j,k)if(f[j][k]<=c&&e[j][k]<=d)g+=1+(c-f[j][k])/e[j][k];d(0,j,k)if(f[j][k]<=d)g+=1+(d-f[j][k])/e[j][k];}cout<<k<<" "<<g%1000000007<<endl;}}