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 int mod=1e9+7;
#define llong long long
int Q;
llong f[110];
llong n,m;
llong a,b;
void calc(int i)
{
	if(!i)return;
	llong ta=a*f[i]+b*f[i-1];
	llong tb=a*f[i+1]+b*f[i];
	a=ta;b=tb;
}
int main()
{
//	freopen("fibo.in","r",stdin);
//	freopen("fibo.out","w",stdout);
	f[0]=f[1]=1;
	for(int i=2;i<=88;++i)f[i]=f[i-1]+f[i-2];
	scanf("%d",&Q);
	int ans;
while(Q--)
{
	scanf("%lld%lld",&n,&m);
	if(n>m)swap(n,m);
	if(n==1)
	{
		printf("%d %lld\n",1,m%mod);
		continue;
	}
	if(n==2&&m==2)
	{
		puts("1 4");
		continue;
	}
	for(int k=87;k>=0;--k)
		if(m>=f[k+1]&&n>=f[k])
		{
			ans=k;
			break;
		}
	printf("%d ",ans);
	llong res=0;
	for(int i=1;i<=ans;++i)
	{
		a=f[i],b=f[i+1];
		calc(ans-i);
		if(a<=n&&b<=m)
		{
			res+=(m-b)/a+1;
			res%=mod;
			if(b<=n)res+=(m-a)/b+1;
			res%=mod;
		}
	}
	printf("%lld\n",res);
}
	return 0;
}