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 L long long
#define pb push_back
using namespace std;
const int q=1000000007;
int n;
L f[100];
vector<L> a[100],b[100];
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k;
	L x,y;
	f[0]=f[1]=1;
	for(i=2;i<=90;i++)
	  f[i]=f[i-1]+f[i-2];
	a[0].pb(0);
	b[0].pb(1);
	a[1].pb(1);
	b[1].pb(2);
	a[1].pb(1);
	b[1].pb(3);
	for(i=1;i<90;i++)
	  for(j=0;j<a[i].size();j++)
	    for(x=b[i][j],y=a[i][j]+b[i][j];y<f[i+3]+f[i];y+=x)
	      {
		   a[i+1].pb(x);
		   b[i+1].pb(y);
		  }
	scanf("%d",&n);
	while(n--)
	  {
	   scanf("%lld%lld",&x,&y);
	   if(x>y)
	     swap(x,y);
	   for(i=2;f[i]<=x && f[i+1]<=y;i++);
	   i--;
	   for(j=0,k=0;j<a[i-1].size();j++)
	     {
	      if(b[i-1][j]<=x && a[i-1][j]+b[i-1][j]<=y)
	        k=(k+(y-a[i-1][j])/b[i-1][j])%q;
	      if(b[i-1][j]<=y && a[i-1][j]+b[i-1][j]<=x)
	        k=(k+(x-a[i-1][j])/b[i-1][j])%q;
		 }
	   if(i==1)
	     k=(k-1+min(x,y)-1)%q;
	   printf("%d %d\n",i,k);
	  }
	return 0;
}