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>
using namespace std;
#define m long long
#define d(s,i,n) for(m i=s;i<n;i++)
#define z else
int main(){m n,a,b,o=1e9+7,x=90,l[x][x],r[x][x];cin>>n;d(0,i,x){l[i][0]=r[i][0]=1;d(1,j,x){if(i==j)r[i][j]=r[i][j-1]*2+l[i][j-1];z r[i][j]=r[i][j-1]+l[i][j-1];l[i][j]=r[i][j-1];}}d(0,i,n){cin>>a>>b;if(b>a)swap(a,b);m k=0;d(0,j,x)if(a>=r[0][j]&&b>=l[0][j])k=j;m g;if(k==0)k=1,g=1;z if(k==1)if(b==1)g=a;z g=4;z{g=1;if(b>=r[0][k]&&a>=l[0][k]&&l[0][k]<r[0][k])g++;d(1,j,k+1)if(r[j][k]<=a&&l[j][k]<=b)g+=1+(a-r[j][k])/l[j][k];d(1,j,k+1)if(r[j][k]<=b)g+=1+(b-r[j][k])/l[j][k];}cout<<k<<" "<<g%o<<" ";}}