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

Solution (in CPP):

Solution:
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
const int p=1000000007;
struct node { LL x,y; } a[102*102];
LL F[102],x,y,X,Y;
int S[102],E[102],i,j,k,n,m,T,g,Z,an;
char c;

inline char getc()
{
	#define VV 10000000
	static char s[VV],*l=s,*r=s;
	if (l==r)
		l=s,r=s+fread(s,1,VV,stdin);
	return l==r?0:*l++;
}

LL read(){ LL z=0; do c=getc(); while (c<'0'||c>'9'); while (c>='0'&&c<='9') z*=10,z+=c-'0',c=getc(); return z; }

bool cmp(node a,node b){ return a.x<b.x; }

int gcd(LL a,LL b)
{
	int z=0;
	if (a<b) swap(a,b);
	while (b)
		a%=b,swap(a,b);
	return z;
}

void solve()
{
	S[1]=1,E[1]=g=2,a[1]=(node){1,2},a[2]=(node){1,3};
	for (i=2;i<=85;++i)
	{
		S[i]=g+1;
		for (j=S[i-1];j<=E[i-1];++j)
		{
			x=a[j].x,y=a[j].y;
			if (x+y<=F[i+2])
				a[++g]=(node){y,x+y};
			if (x+(y<<1)<=F[i+2])
				a[++g]=(node){y,x+(y<<1)};
		}
		sort(a+S[i],a+g+1,cmp);
		k=S[i];
		for (j=S[i]+1;j<=g;++j)
			if (a[k].x!=a[j].x||a[k].y!=a[j].y)
				a[++k]=a[j];
		E[i]=k;
	}
}

int main()
{
	T=read();
	for (F[0]=F[1]=1,i=2;i<=90;++i) F[i]=F[i-1]+F[i-2];
	solve();
	while (T--)
	{
		an=0,X=read(),Y=read();
		if (X>Y) swap(X,Y);
		
		for (Z=2;F[Z]<=X&&F[Z+1]<=Y;++Z);
		--Z;
		printf("%d ",Z);
		if (Z==1)
			printf("%d\n",X%p*(Y%p)%p);
		else
		{
			for (i=S[Z-1];i<=E[Z-1];++i)
			{
				x=a[i].x,y=a[i].y;
				if (y<=X)
					an=(an+(Y-x)/y)%p;
				if (y<=Y)
					an=(an+(X-x)/y)%p;
			}
			printf("%d\n",an);
		}
	}
	return 0;
}