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>
using namespace std;
const long long inf=1e18;
const long long Mod=1000000007;
struct node
{
	long long x,y;
};
int q;
long long fib[100];
vector<node> v[100],ch;
int main()
{
	fib[0]=fib[1]=1;
	int k=1;
	while (true)
	{
		k++;
		fib[k]=fib[k-1]+fib[k-2];
		if (fib[k]>inf)
		{
			break;
		}
	}
	k--;
	v[1].push_back((node){1,2});
	v[1].push_back((node){1,3});
	v[1].push_back((node){1,4});
	for (int i=1;i<k-2;i++)
	{
		for (int j=0;j<v[i].size();j++)
		{
			long long x=v[i][j].x;
			long long y=v[i][j].y;
			x+=y;
			while (x<=fib[i]+fib[i+3])
			{
				v[i+1].push_back((node){y,x});
				x+=y;
			}
		}
	}
	scanf("%d",&q);
	while (q--)
	{
		long long n,m;
		scanf("%lld%lld",&n,&m);
		if (n>m)
		{
			swap(n,m);
		}
		k=1;
		while (fib[k+1]<=n && fib[k+2]<=m)
		{
			k++;
		}
		printf("%d ",k);
		if (k==1)
		{
			n%=Mod;
			m%=Mod;
			printf("%lld\n",n*m%Mod);
			continue;
		}
		long long ans=0;
		k--;
		for (int i=0;i<v[k].size();i++)
		{
			long long x=v[k][i].x;
			long long y=v[k][i].y;
			if (y<=n)
			{
				ans=(ans+(m-x)/y)%Mod;
			}
			if (y<=m)
			{
				ans=(ans+(n-x)/y)%Mod;
			}
		}
		printf("%lld\n",ans);
	}
    return 0;
}
//凝眸顾盼柔情藏，灵动佳人婉清扬。