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<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<ctime>
#include<map>
#include<bitset>
#include<set>
#define LL long long
#define mp(x,y) make_pair(x,y)
#define pll pair<long long,long long>
#define pii pair<int,int>
using namespace std;
inline LL read()
{
	LL f=1,x=0;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int stack[20];
template<typename T>inline void write(T x)
{
	if(x<0){putchar('-');x=-x;}
    if(!x){putchar('0');return;}
    int top=0;
    while(x)stack[++top]=x%10,x/=10;
    while(top)putchar(stack[top--]+'0');
}
template<typename T>inline void pr1(T x){write(x);putchar(' ');}
template<typename T>inline void pr2(T x){write(x);putchar('\n');}
const int MAXN=105;
const int mod=1e9+7;
vector<pll> vec[MAXN];
LL f[MAXN],tot;
void ad(int &x,int y){x+=y;if(x>=mod)x-=mod;}
int main()
{
	f[0]=f[1]=1;tot=1;
	while(f[tot]<1e18)
	{
		f[tot+1]=f[tot]+f[tot-1];
		++tot;
	}
	vec[1].push_back(mp(1,2));vec[1].push_back(mp(1,3));vec[1].push_back(mp(1,4));
	for(int i=2;i<=tot;i++)
		for(int j=0;j<vec[i-1].size();j++)
		{
			LL x=vec[i-1][j].first,y=vec[i-1][j].second;
			x+=y;
			while(x<=f[i-1]+f[i+2])
			{
				vec[i].push_back(mp(y,x));
				x+=y;
			}
		}
	int T=read();while(T--)
	{
		LL n=read(),m=read();
		if(n>m)swap(n,m);int ans=1;
		while(f[ans+1]<=n&&f[ans+2]<=m)++ans;
		pr1(ans);
		if(ans==1){pr2(1LL*n*m%mod);continue;}
		int sum=0;
		for(int i=0;i<vec[ans-1].size();i++)
		{
			LL x=vec[ans-1][i].first,y=vec[ans-1][i].second;
			if(y<=n)ad(sum,1LL*(m-x)/y%mod);
			if(y<=m)ad(sum,1LL*(n-x)/y%mod);
		}
		pr2(sum);
	}
	return 0;
}