#include<bits/stdc++.h>
using namespace std;
long long fib[233],x,y,ans;
int now;long long all;
void dfs(long long u,long long v,int c,int err)
{
	//cout<<u<<' '<<v<<' '<<c<<' '<<err<<' '<<(((u>=fib[c+2])&&(v>=fib[c+3]))||(u>x)||(v>y))<<endl;
	if (((u>=fib[c+2])&&(v>=fib[c+3]))||(u>x)||(v>y)) return;
	//if (c==now) {ans+=(int)((u<=x)&&(v<=y))+(int)((v<=x)&&(u<=y));return;}
	if (c==now-1)
	{
		ans+=max(0ll,(y-u)/v)*(v<=x)+max(0ll,(x-u)/v)*(v<=y);
		return;
	}
	if (!err)
	{
		dfs(v,u+v,c+1,0);
		dfs(v,u+v+v,c+1,1);
		//v,u+kv
		//p=fib[now-c-2]*v+fib[now-c-1]*(u+kv)
		//q=fib[now-c-1]*v+fib[now-c]*(u+kv)
		//p<=x q<=y
	}
	else if (err==1)
	{
		if ((double)u*fib[now-2-c]+(double)v*fib[now-1-c]>1.1e18) return;
		if ((double)u*fib[now-1-c]+(double)v*fib[now-0-c]>1.1e18) return;
		long long uu=u*fib[now-2-c]+v*fib[now-1-c],vv=u*fib[now-1-c]+v*fib[now-0-c];
		dfs(uu,vv,now-1,1);
	}
}
int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	fib[0]=0;fib[1]=1;for (int i=2;i<=90;i++) fib[i]=fib[i-1]+fib[i-2];
	int q;scanf("%d",&q);
	while (q--)
	{//if (q%1000==0) cerr<<q<<endl;
		scanf("%lld%lld",&x,&y);if (x>y) swap(x,y);
		if (x==1) {printf("1 %lld\n",y%1000000007);continue;}
		if (y==2) {printf("1 4\n");continue;}
		now=0;while ((fib[now+2]<=x)&&(fib[now+3]<=y)) now++;
		printf("%d ",now);
		ans=0;dfs(1,1,0,0);
		printf("%lld\n",ans%1000000007);
	}
	cerr<<all<<endl;
}
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;
long long fib[233],x,y,ans;
int now;long long all;
void dfs(long long u,long long v,int c,int err)
{
	//cout<<u<<' '<<v<<' '<<c<<' '<<err<<' '<<(((u>=fib[c+2])&&(v>=fib[c+3]))||(u>x)||(v>y))<<endl;
	if (((u>=fib[c+2])&&(v>=fib[c+3]))||(u>x)||(v>y)) return;
	//if (c==now) {ans+=(int)((u<=x)&&(v<=y))+(int)((v<=x)&&(u<=y));return;}
	if (c==now-1)
	{
		ans+=max(0ll,(y-u)/v)*(v<=x)+max(0ll,(x-u)/v)*(v<=y);
		return;
	}
	if (!err)
	{
		dfs(v,u+v,c+1,0);
		dfs(v,u+v+v,c+1,1);
		//v,u+kv
		//p=fib[now-c-2]*v+fib[now-c-1]*(u+kv)
		//q=fib[now-c-1]*v+fib[now-c]*(u+kv)
		//p<=x q<=y
	}
	else if (err==1)
	{
		if ((double)u*fib[now-2-c]+(double)v*fib[now-1-c]>1.1e18) return;
		if ((double)u*fib[now-1-c]+(double)v*fib[now-0-c]>1.1e18) return;
		long long uu=u*fib[now-2-c]+v*fib[now-1-c],vv=u*fib[now-1-c]+v*fib[now-0-c];
		dfs(uu,vv,now-1,1);
	}
}
int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	fib[0]=0;fib[1]=1;for (int i=2;i<=90;i++) fib[i]=fib[i-1]+fib[i-2];
	int q;scanf("%d",&q);
	while (q--)
	{//if (q%1000==0) cerr<<q<<endl;
		scanf("%lld%lld",&x,&y);if (x>y) swap(x,y);
		if (x==1) {printf("1 %lld\n",y%1000000007);continue;}
		if (y==2) {printf("1 4\n");continue;}
		now=0;while ((fib[now+2]<=x)&&(fib[now+3]<=y)) now++;
		printf("%d ",now);
		ans=0;dfs(1,1,0,0);
		printf("%lld\n",ans%1000000007);
	}
	cerr<<all<<endl;
}
```