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>
#include<cstdio>
#include<vector>
using namespace std;
#define ll long long
#define MOD 1000000007
#define pi pair<ll,ll>
#define mp make_pair
inline ll read()
{
	ll x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
ll f[110];
vector<pi> a[110];
int main()
{
	int T=read();
	f[0]=f[1]=1;
	for(int i=2;i<=100;++i)f[i]=f[i-1]+f[i-2];
	a[1].push_back(mp(1,2));a[1].push_back(mp(1,3));a[1].push_back(mp(1,4));
	for(int i=1;i<=100;++i)
		for(int j=0,l=a[i].size();j<l;++j)
		{
			ll x=a[i][j].second,y=a[i][j].first+x;
			while(y<=f[i+3]+f[i])a[i+1].push_back(mp(x,y)),y+=x;
		}
	while(T--)
	{
		ll n=read(),m=read();if(n>m)swap(n,m);
		if(n==1&&m==1){puts("1 1");continue;}
		int p,ans=0;for(int i=1;;++i)if(!(n>=f[i]&&m>=f[i+1])){p=i-1;break;}
		printf("%d ",p);if(p==1){printf("%lld\n",(n%MOD)*(m%MOD)%MOD);continue;}
		for(int j=0,l=a[p-1].size();j<l;++j)
		{
			ll x=a[p-1][j].first,y=a[p-1][j].second;
			if(y<=n)ans=(ans+(m-x)/y)%MOD;
			if(y<=m)ans=(ans+(n-x)/y)%MOD;
		}
		printf("%d\n",ans);
	}
}