#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define LL long long
#define N 500
#define mmod 1000000007
using namespace std;
LL f[N],cnt[N],a[N][N][2],x,y,ans;
int n;
void solve1(LL x,LL y)
{
	for(int i=1;i<=x;i++)
	{
		ans=(ans+y/i)%mmod;
		if(y/i==0) break;
	}
	for(int i=1;i<=y;i++)
	{
		ans=(ans+x/i)%mmod;
		if(x/i==0) break;
	}
	ans=(ans-min(x,y))%mmod;
	ans=(ans+mmod)%mmod;
}
void solve(LL a,LL b,LL x,LL y)
{
	if(b>x || b+a>y) return;
	ans=(ans+(y-a)/b)%mmod;
}
int main()
{
	f[0]=f[1]=1;
	for(int i=2;;i++)
	{
		f[i]=f[i-1]+f[i-2];
		if(f[i]>1e18) {n=i;break;}
	}
	n-=2;
	cnt[1]=2;
	a[1][1][0]=1;a[1][1][1]=2;a[1][2][0]=1;a[1][2][1]=3;
	for(int i=1;i<n;i++)
		for(int j=1;j<=cnt[i];j++)
		{
			LL t=a[i][j][0]+a[i][j][1];
			while(t<=f[i+3])
			{
				a[i+1][++cnt[i+1]][0]=a[i][j][1];
				a[i+1][cnt[i+1]][1]=t;
				t+=a[i][j][1];
			}
		}
	int z;scanf("%d",&z);
	while(z--)
	{
		scanf("%lld%lld",&x,&y);
		if(x>y) swap(x,y);
		int k=1;
		for(int i=1;i<=n;i++) 
			if(f[i]<=x && f[i+1]<=y) k=i;
			else break;
		ans=0;
		printf("%d ",k);
		if(k==1) solve1(x,y);
		else for(int i=1;i<=cnt[k-1];i++) solve(a[k-1][i][0],a[k-1][i][1],x,y),solve(a[k-1][i][0],a[k-1][i][1],y,x);
		printf("%lld\n",ans);
	}
	return 0;
}
### Prompt

Your challenge is to write 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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define LL long long
#define N 500
#define mmod 1000000007
using namespace std;
LL f[N],cnt[N],a[N][N][2],x,y,ans;
int n;
void solve1(LL x,LL y)
{
	for(int i=1;i<=x;i++)
	{
		ans=(ans+y/i)%mmod;
		if(y/i==0) break;
	}
	for(int i=1;i<=y;i++)
	{
		ans=(ans+x/i)%mmod;
		if(x/i==0) break;
	}
	ans=(ans-min(x,y))%mmod;
	ans=(ans+mmod)%mmod;
}
void solve(LL a,LL b,LL x,LL y)
{
	if(b>x || b+a>y) return;
	ans=(ans+(y-a)/b)%mmod;
}
int main()
{
	f[0]=f[1]=1;
	for(int i=2;;i++)
	{
		f[i]=f[i-1]+f[i-2];
		if(f[i]>1e18) {n=i;break;}
	}
	n-=2;
	cnt[1]=2;
	a[1][1][0]=1;a[1][1][1]=2;a[1][2][0]=1;a[1][2][1]=3;
	for(int i=1;i<n;i++)
		for(int j=1;j<=cnt[i];j++)
		{
			LL t=a[i][j][0]+a[i][j][1];
			while(t<=f[i+3])
			{
				a[i+1][++cnt[i+1]][0]=a[i][j][1];
				a[i+1][cnt[i+1]][1]=t;
				t+=a[i][j][1];
			}
		}
	int z;scanf("%d",&z);
	while(z--)
	{
		scanf("%lld%lld",&x,&y);
		if(x>y) swap(x,y);
		int k=1;
		for(int i=1;i<=n;i++) 
			if(f[i]<=x && f[i+1]<=y) k=i;
			else break;
		ans=0;
		printf("%d ",k);
		if(k==1) solve1(x,y);
		else for(int i=1;i<=cnt[k-1];i++) solve(a[k-1][i][0],a[k-1][i][1],x,y),solve(a[k-1][i][0],a[k-1][i][1],y,x);
		printf("%lld\n",ans);
	}
	return 0;
}
```