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 l int64_t
#define c continue;
#define o(a) cout<<a<<endl;
#define r(i,x,y) for(l i=x;i<y;i++){
#define t(a,b,c) t[a][b][c]
using namespace std;
int main(){l m=1e9+7,x,y,z,w,o,p,q,v[93],t[91][91][2],a,b;v[0]=1,v[1]=1;r(i,2,93)v[i]=v[i-1]+v[i-2];}r(i,2,91)r(j,1,i)a=v[i],b=v[i+1];if(j<2){t(i,0,0)=a,t(i,0,1)=b,t(i,j,0)=b,t(i,j,1)=2*b-a;c;}t(i,j,0)=t(i-1,j-1,1),t(i,j,1)=t(i-1,j-1,1)+t(i-1,j-1,0);}}cin>>q;while(q--){cin>>x>>y;if(x>y)swap(x,y);if(y<2){o("1 1")c}r(i,0,93)if(v[i]>x||v[i+1]>y){z=i-1;break;}}o(z)if(z<2){o(x*y%m)c}w=0;r(j,0,z)o=t(z,j,0),p=t(z,j,1);if(o>x||p>y)c w+=(y-p)/o+1;if(x>=p)w+=(x-p)/o+1;}o(w%m)}}