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
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;for(l i=2;i<93;i++)v[i]=v[i-1]+v[i-2];for(l i=2;i<91;i++){for(l j=1;j<i;j++){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;}for(l i=0;i<93;i++){if(v[i]>x||v[i+1]>y){z=i-1;break;}}o(z);if(z<2){o(x*y%m);c;}w=0;for(l j=0;j<z;j++){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);}}