#include<bits/stdc++.h>

typedef long long ll;
typedef unsigned long long ull;

using namespace std;

const int MAXN=233;
const int P=1E9+7;
const ll INF=1E18;

template<typename T1,typename T2>
	void Inc(T1 &a,T2 b){a+=b;if(a>=P) a-=P;}

template<typename T1,typename T2>
	void Dec(T1 &a,T2 b){a-=b;if(a<0) a+=P;}

template<typename T1,typename T2>
	T1 Add(T1 a,T2 b){a+=b;return a>=P?a-P:a;}

template<typename T1,typename T2>
	T1 Sub(T1 a,T2 b){a-=b;return a<0?a+P:a;}

int times(ll a,ll b)
{
	if(a<b) swap(a,b);
	if(b==0) return 0;
	return times(b,a%b)+1;
}

int q,LIM;
ll fib[MAXN];
vector<pair<ll,ll> > ep[MAXN];

int main()
{
	fib[0]=fib[1]=1;LIM=1;
	for(int i=2;fib[i-1]<=INF;i++)
		fib[i]=fib[i-1]+fib[i-2],++LIM;
	ep[1].emplace_back(1,2);
	ep[1].emplace_back(1,3);
	for(int i=1;i<LIM;i++)
	{
		for(auto x:ep[i])
		{
			ll t=x.first+x.second;
			while(t<=fib[i+3])
			{
				if(times(x.second,t)==i+1)
					ep[i+1].emplace_back(x.second,t);
				t+=x.second;
			}
		}
		sort(ep[i+1].begin(),ep[i+1].end());
		ep[i+1].erase(unique(ep[i+1].begin(),ep[i+1].end()),ep[i+1].end());
	}
	scanf("%d",&q);
	while(q--)
	{
		ll A,B,Ans1=1,Ans2=0;
		scanf("%lld%lld",&A,&B);
		if(A>B) swap(A,B);
		while(fib[Ans1+1]<=A&&fib[Ans1+2]<=B)
			++Ans1,assert(Ans1<=LIM);
		if(Ans1==1) printf("1 %lld\n",1ll*A*B%P);
		else
		{
			for(auto x:ep[Ans1-1])
			{
				if(A>=x.second)
					Inc(Ans2,(B-x.first)/x.second%P);
				if(B>=x.second)
					Inc(Ans2,(A-x.first)/x.second%P);
			}
			printf("%lld %lld\n",Ans1,Ans2);
		}
	}
}

### Prompt

Create a solution in Cpp for 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<bits/stdc++.h>

typedef long long ll;
typedef unsigned long long ull;

using namespace std;

const int MAXN=233;
const int P=1E9+7;
const ll INF=1E18;

template<typename T1,typename T2>
	void Inc(T1 &a,T2 b){a+=b;if(a>=P) a-=P;}

template<typename T1,typename T2>
	void Dec(T1 &a,T2 b){a-=b;if(a<0) a+=P;}

template<typename T1,typename T2>
	T1 Add(T1 a,T2 b){a+=b;return a>=P?a-P:a;}

template<typename T1,typename T2>
	T1 Sub(T1 a,T2 b){a-=b;return a<0?a+P:a;}

int times(ll a,ll b)
{
	if(a<b) swap(a,b);
	if(b==0) return 0;
	return times(b,a%b)+1;
}

int q,LIM;
ll fib[MAXN];
vector<pair<ll,ll> > ep[MAXN];

int main()
{
	fib[0]=fib[1]=1;LIM=1;
	for(int i=2;fib[i-1]<=INF;i++)
		fib[i]=fib[i-1]+fib[i-2],++LIM;
	ep[1].emplace_back(1,2);
	ep[1].emplace_back(1,3);
	for(int i=1;i<LIM;i++)
	{
		for(auto x:ep[i])
		{
			ll t=x.first+x.second;
			while(t<=fib[i+3])
			{
				if(times(x.second,t)==i+1)
					ep[i+1].emplace_back(x.second,t);
				t+=x.second;
			}
		}
		sort(ep[i+1].begin(),ep[i+1].end());
		ep[i+1].erase(unique(ep[i+1].begin(),ep[i+1].end()),ep[i+1].end());
	}
	scanf("%d",&q);
	while(q--)
	{
		ll A,B,Ans1=1,Ans2=0;
		scanf("%lld%lld",&A,&B);
		if(A>B) swap(A,B);
		while(fib[Ans1+1]<=A&&fib[Ans1+2]<=B)
			++Ans1,assert(Ans1<=LIM);
		if(Ans1==1) printf("1 %lld\n",1ll*A*B%P);
		else
		{
			for(auto x:ep[Ans1-1])
			{
				if(A>=x.second)
					Inc(Ans2,(B-x.first)/x.second%P);
				if(B>=x.second)
					Inc(Ans2,(A-x.first)/x.second%P);
			}
			printf("%lld %lld\n",Ans1,Ans2);
		}
	}
}

```