#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<cmath>
#include<functional>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
void sort(int &a,int &b)
{
	if(a>b)
		swap(a,b);
}
void open(const char *s)
{
#ifndef ONLINE_JUDGE
	char str[100];
	sprintf(str,"%s.in",s);
	freopen(str,"r",stdin);
	sprintf(str,"%s.out",s);
	freopen(str,"w",stdout);
#endif
}
int rd()
{
	int s=0,c;
	while((c=getchar())<'0'||c>'9');
	do
	{
		s=s*10+c-'0';
	}
	while((c=getchar())>='0'&&c<='9');
	return s;
}
int upmin(int &a,int b)
{
	if(b<a)
	{
		a=b;
		return 1;
	}
	return 0;
}
int upmax(int &a,int b)
{
	if(b>a)
	{
		a=b;
		return 1;
	}
	return 0;
}
vector<pll> a[100];
ll inf=1000000000000000000ll;
ll f[100];
ll p=1000000007;
void init()
{
	f[0]=f[1]=1;
	int n=1;
	while(1)
	{
		f[n+1]=f[n-1]+f[n];
		if(f[n+1]>inf)
			break;
		n++;
	}
	a[1].push_back(pll(1,2));
	a[1].push_back(pll(1,3));
	a[1].push_back(pll(1,4));
	int i;
	ll x,y;
	for(i=1;i<=n-3;i++)
		for(auto v:a[i])
		{
			x=v.first;
			y=v.second;
			for(x+=y;x<=f[i+3]+f[i];x+=y)
				a[i+1].push_back(pll(y,x));
		}
}
int main()
{
	init();
	int t;
	ll n,m,x,y;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld",&n,&m);
		if(n>m)
			swap(n,m);
		int s=1;
		while(f[s+1]<=n&&f[s+2]<=m)
			s++;
		printf("%d ",s);
		if(s==1)
			printf("%lld\n",n*m%p);
		else
		{
			ll ans=0;
			for(auto v:a[s-1])
			{
				x=v.first;
				y=v.second;
				if(y<=n)
					ans+=(m-x)/y%p;
				if(y<=m)
					ans+=(n-x)/y%p;
				ans%=p;
			}
			printf("%lld\n",ans);
		}
	}
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<cmath>
#include<functional>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
void sort(int &a,int &b)
{
	if(a>b)
		swap(a,b);
}
void open(const char *s)
{
#ifndef ONLINE_JUDGE
	char str[100];
	sprintf(str,"%s.in",s);
	freopen(str,"r",stdin);
	sprintf(str,"%s.out",s);
	freopen(str,"w",stdout);
#endif
}
int rd()
{
	int s=0,c;
	while((c=getchar())<'0'||c>'9');
	do
	{
		s=s*10+c-'0';
	}
	while((c=getchar())>='0'&&c<='9');
	return s;
}
int upmin(int &a,int b)
{
	if(b<a)
	{
		a=b;
		return 1;
	}
	return 0;
}
int upmax(int &a,int b)
{
	if(b>a)
	{
		a=b;
		return 1;
	}
	return 0;
}
vector<pll> a[100];
ll inf=1000000000000000000ll;
ll f[100];
ll p=1000000007;
void init()
{
	f[0]=f[1]=1;
	int n=1;
	while(1)
	{
		f[n+1]=f[n-1]+f[n];
		if(f[n+1]>inf)
			break;
		n++;
	}
	a[1].push_back(pll(1,2));
	a[1].push_back(pll(1,3));
	a[1].push_back(pll(1,4));
	int i;
	ll x,y;
	for(i=1;i<=n-3;i++)
		for(auto v:a[i])
		{
			x=v.first;
			y=v.second;
			for(x+=y;x<=f[i+3]+f[i];x+=y)
				a[i+1].push_back(pll(y,x));
		}
}
int main()
{
	init();
	int t;
	ll n,m,x,y;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld",&n,&m);
		if(n>m)
			swap(n,m);
		int s=1;
		while(f[s+1]<=n&&f[s+2]<=m)
			s++;
		printf("%d ",s);
		if(s==1)
			printf("%lld\n",n*m%p);
		else
		{
			ll ans=0;
			for(auto v:a[s-1])
			{
				x=v.first;
				y=v.second;
				if(y<=n)
					ans+=(m-x)/y%p;
				if(y<=m)
					ans+=(n-x)/y%p;
				ans%=p;
			}
			printf("%lld\n",ans);
		}
	}
	return 0;
}
```