/*
srt: 2017年07月04日 15時22分28秒
鹧鸪天·晚日寒鸦一片愁 辛弃疾
晚日寒鸦一片愁。柳塘新绿却温柔。若教眼底无离恨，不信人间有白头。
肠已断，泪难收。相思重上小红楼。情知已被山遮断，频倚阑干不自由。 
end: 2017年07月04日 16時39分11秒
*/

#include<cstdio>
#include<algorithm>

typedef long long ll;
const int mod=1e9+7;
ll f[100];int n;
// 1,2,3,5,8
// x,y,x+y,x+2y,2x+3y,3x+5y,5x+8y

void forward(ll x,ll y,int i,ll&c,ll lima,ll limb)
{
	ll a,b;
	if(i<2)
	{
		if(i==0)a=x,b=y;
		else a=y,b=x+y;
		if(a>lima||b>limb)return;
	}else{
		if(lima/x<f[i-2]||lima/y<f[i-1]||x*f[i-2]>lima-y*f[i-1])return;
		if(limb/x<f[i-1]||limb/y<f[ i ]||x*f[i-1]>limb-y*f[ i ])return;
		a=x*f[i-2]+y*f[i-1];
		b=x*f[i-1]+y*f[i];
	}
	(c+=(limb-b)/a+1)%=mod;
	if(b<=lima)
		(c+=(lima-b)/a+1)%=mod;
}

void work(ll x,ll y)
{
	int i,k=std::upper_bound(f+1,f+n,x)-f-1;
	if(f[k+1]>y)k=std::upper_bound(f+1,f+n,y)-f-2;
	ll c=((y-f[k+1])/f[k]+1)%mod;
	if(f[k+1]<=x)
		(c+=(x-f[k+1])/f[k]+1)%=mod;
	for(i=1;i<k;i++)
	{
		forward(f[i],2*f[i]+f[i-1],k-i,c,x,y);
		forward(f[i],3*f[i]+f[i-1],k-i,c,x,y);
	}
	printf("%d %lld\n",k,c);
}

int main()
{
	ll x,y;int i;
	f[0]=f[1]=1;
	for(n=1;f[n]>0;n++)
		f[n+1]=f[n]+f[n-1];
	scanf("%d",&i);
	while(i--)
	{
		scanf("%lld%lld\n",&x,&y);
		if(x==1||y==1||(x==2&&y==2))
			printf("1 %lld\n",x*y%mod);
		else
			x>y&&(x^=y^=x^=y),work(x,y);
	}
}
### Prompt

Please provide a cpp coded solution 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
/*
srt: 2017年07月04日 15時22分28秒
鹧鸪天·晚日寒鸦一片愁 辛弃疾
晚日寒鸦一片愁。柳塘新绿却温柔。若教眼底无离恨，不信人间有白头。
肠已断，泪难收。相思重上小红楼。情知已被山遮断，频倚阑干不自由。 
end: 2017年07月04日 16時39分11秒
*/

#include<cstdio>
#include<algorithm>

typedef long long ll;
const int mod=1e9+7;
ll f[100];int n;
// 1,2,3,5,8
// x,y,x+y,x+2y,2x+3y,3x+5y,5x+8y

void forward(ll x,ll y,int i,ll&c,ll lima,ll limb)
{
	ll a,b;
	if(i<2)
	{
		if(i==0)a=x,b=y;
		else a=y,b=x+y;
		if(a>lima||b>limb)return;
	}else{
		if(lima/x<f[i-2]||lima/y<f[i-1]||x*f[i-2]>lima-y*f[i-1])return;
		if(limb/x<f[i-1]||limb/y<f[ i ]||x*f[i-1]>limb-y*f[ i ])return;
		a=x*f[i-2]+y*f[i-1];
		b=x*f[i-1]+y*f[i];
	}
	(c+=(limb-b)/a+1)%=mod;
	if(b<=lima)
		(c+=(lima-b)/a+1)%=mod;
}

void work(ll x,ll y)
{
	int i,k=std::upper_bound(f+1,f+n,x)-f-1;
	if(f[k+1]>y)k=std::upper_bound(f+1,f+n,y)-f-2;
	ll c=((y-f[k+1])/f[k]+1)%mod;
	if(f[k+1]<=x)
		(c+=(x-f[k+1])/f[k]+1)%=mod;
	for(i=1;i<k;i++)
	{
		forward(f[i],2*f[i]+f[i-1],k-i,c,x,y);
		forward(f[i],3*f[i]+f[i-1],k-i,c,x,y);
	}
	printf("%d %lld\n",k,c);
}

int main()
{
	ll x,y;int i;
	f[0]=f[1]=1;
	for(n=1;f[n]>0;n++)
		f[n+1]=f[n]+f[n-1];
	scanf("%d",&i);
	while(i--)
	{
		scanf("%lld%lld\n",&x,&y);
		if(x==1||y==1||(x==2&&y==2))
			printf("1 %lld\n",x*y%mod);
		else
			x>y&&(x^=y^=x^=y),work(x,y);
	}
}
```