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<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define pll pair<ll,ll>
#define itt vector<pll>::iterator
#define inf 2333333333333333333LL
#define mod 1000000007LL

vector<pll> v[233];

ll fib[233],t,n,m,ans;

inline ll rd()
{
	ll x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

int main()
{
	t=rd();fib[0]=fib[1]=1;
	for (ll i=2;i<=88;i++) fib[i]=fib[i-1]+fib[i-2];
	v[1].push_back(make_pair(1,2));
	v[1].push_back(make_pair(1,3));
	v[1].push_back(make_pair(1,4));
	for (ll i=1;i<85;i++)
	{
		for (itt it=v[i].begin();it!=v[i].end();it++)
		{
			ll x=(*it).first,y=(*it).second;
			x+=y;
			while (x<=fib[i+3]+fib[i])
			{
				v[i+1].push_back(make_pair(y,x));
				x+=y;
			}
		}
	}
	while (t--)
	{
		n=rd();m=rd();
		if (n>m) swap(n,m);
		ans=1;
		while (fib[ans+1]<=n&&fib[ans+2]<=m) ans++;
		printf("%lld ",ans);
		if (ans==1) { printf("%lld\n",n*m%mod);continue; }
		ll res=0;
		for (itt it=v[ans-1].begin();it!=v[ans-1].end();it++)
		{
			ll x=(*it).first,y=(*it).second;
			if (y<=n) (res+=(m-x)/y)%=mod;
			if (y<=m) (res+=(n-x)/y)%=mod;
		}
		printf("%lld\n",res);
	}
	return 0;
}