#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

#define fir first
#define sec second

typedef long long LL;
typedef pair<LL,LL> PP;

namespace INPUT{
	const int L=1<<15;
	char _buf[L],*S,*T,c;
	char _gc(){
		if(S==T){
			T=(S=_buf)+fread(_buf,1,L,stdin);
			if(S==T) return EOF;
		}
		return *S++;
	}
	void readi(int &X){
		for(c=_gc();c<'0'||c>'9';c=_gc());X=c&15;
		for(c=_gc();c>='0'&&c<='9';X=X*10+(c&15),c=_gc());
	}
	void readl(LL &X){
		for(c=_gc();c<'0'||c>'9';c=_gc());X=c&15;
		for(c=_gc();c>='0'&&c<='9';X=X*10+(c&15),c=_gc());
	}
}
using INPUT::readi;
using INPUT::readl;

const int Maxn=100;
const LL Maxx=(LL)1E18;
const int Mod=1E9+7;

int Q;
LL X,Y;
LL fib[Maxn];
PP s[Maxn][Maxn];

LL GCD(LL X,LL Y){
	if(X<Y) swap(X,Y);
	if(Y==0) return 0;
	return GCD(Y,X%Y)+1;
}

int main(){
	fib[1]=1,fib[2]=2;
	for(int i=3;;++i){
		fib[i]=fib[i-1]+fib[i-2];
		if(fib[i]>Maxx) break;
	}
	s[2][1]=PP(2,3),s[2][2]=PP(3,4);
	for(int i=3;;++i){
		if(fib[i]>Maxx) break;
		for(int j=1;j<i;++j)
			s[i][j]=PP(s[i-1][j].sec,s[i-1][j].fir+s[i-1][j].sec);
		s[i][i]=PP(fib[i+1],fib[i-1]+fib[i+1]);
	}
	//for(int i=2;i<=5;++i,cerr<<endl)
	//	for(int j=1;j<=i;++j) cerr<<s[i][j].fir<<' '<<s[i][j].sec<<endl;
	readi(Q);
	while(Q--){
		LL Ans=0;
		readl(X),readl(Y);
		if(X==1&&Y==1){puts("1 1");continue;}
		if(X>Y) swap(X,Y);
		int t=1;
		for(t=1;;++t)
			if(fib[t]>X||fib[t+1]>Y) break;
		--t;
		if(t==1){
			printf("%d %lld\n",t,X*Y%Mod);
			continue ;
		}
		for(int i=1;i<=t;++i)
			if(s[t][i].fir<=X&&s[t][i].sec<=Y)
				(Ans+=(Y-s[t][i].sec)/s[t][i].fir+1)%=Mod;
		for(int i=1;i<=t;++i)
			if(s[t][i].fir<=Y&&s[t][i].sec<=X)
				(Ans+=(X-s[t][i].sec)/s[t][i].fir+1)%=Mod;
		printf("%d %lld\n",t,Ans);
	}
	return 0;
}

### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

#define fir first
#define sec second

typedef long long LL;
typedef pair<LL,LL> PP;

namespace INPUT{
	const int L=1<<15;
	char _buf[L],*S,*T,c;
	char _gc(){
		if(S==T){
			T=(S=_buf)+fread(_buf,1,L,stdin);
			if(S==T) return EOF;
		}
		return *S++;
	}
	void readi(int &X){
		for(c=_gc();c<'0'||c>'9';c=_gc());X=c&15;
		for(c=_gc();c>='0'&&c<='9';X=X*10+(c&15),c=_gc());
	}
	void readl(LL &X){
		for(c=_gc();c<'0'||c>'9';c=_gc());X=c&15;
		for(c=_gc();c>='0'&&c<='9';X=X*10+(c&15),c=_gc());
	}
}
using INPUT::readi;
using INPUT::readl;

const int Maxn=100;
const LL Maxx=(LL)1E18;
const int Mod=1E9+7;

int Q;
LL X,Y;
LL fib[Maxn];
PP s[Maxn][Maxn];

LL GCD(LL X,LL Y){
	if(X<Y) swap(X,Y);
	if(Y==0) return 0;
	return GCD(Y,X%Y)+1;
}

int main(){
	fib[1]=1,fib[2]=2;
	for(int i=3;;++i){
		fib[i]=fib[i-1]+fib[i-2];
		if(fib[i]>Maxx) break;
	}
	s[2][1]=PP(2,3),s[2][2]=PP(3,4);
	for(int i=3;;++i){
		if(fib[i]>Maxx) break;
		for(int j=1;j<i;++j)
			s[i][j]=PP(s[i-1][j].sec,s[i-1][j].fir+s[i-1][j].sec);
		s[i][i]=PP(fib[i+1],fib[i-1]+fib[i+1]);
	}
	//for(int i=2;i<=5;++i,cerr<<endl)
	//	for(int j=1;j<=i;++j) cerr<<s[i][j].fir<<' '<<s[i][j].sec<<endl;
	readi(Q);
	while(Q--){
		LL Ans=0;
		readl(X),readl(Y);
		if(X==1&&Y==1){puts("1 1");continue;}
		if(X>Y) swap(X,Y);
		int t=1;
		for(t=1;;++t)
			if(fib[t]>X||fib[t+1]>Y) break;
		--t;
		if(t==1){
			printf("%d %lld\n",t,X*Y%Mod);
			continue ;
		}
		for(int i=1;i<=t;++i)
			if(s[t][i].fir<=X&&s[t][i].sec<=Y)
				(Ans+=(Y-s[t][i].sec)/s[t][i].fir+1)%=Mod;
		for(int i=1;i<=t;++i)
			if(s[t][i].fir<=Y&&s[t][i].sec<=X)
				(Ans+=(X-s[t][i].sec)/s[t][i].fir+1)%=Mod;
		printf("%d %lld\n",t,Ans);
	}
	return 0;
}

```