#include<bits/stdc++.h>
using namespace std;
#define reg register
#define ll long long
#define mo 1000000007
#define P make_pair
inline ll read()
{
	ll x=0,w=0;char ch=getchar();
	while(!isdigit(ch))w|=ch=='-',ch=getchar();
	while(isdigit(ch))x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return w?-x:x;
}
inline void Max(int &x,int y){x=x>y?x:y;}
int q,ans,res;
ll f[100],x,y;
vector< pair<ll,ll> >G[100];
inline void Add(int &x,int y){x+=y;if(x>=mo)x-=mo;}
int main()
{
    q=read();reg int i,o,sz,b;
    f[0]=1;f[1]=2;reg ll xx,yy;
    for(i=2;i<=89;++i)f[i]=f[i-1]+f[i-2];
    G[1].push_back(P(1,2));G[1].push_back(P(1,3));G[1].push_back(P(1,4));
    for(i=1;i<87;++i)
    {
        sz=G[i].size();
        for(o=0;o<sz;++o)
        {
            x=G[i][o].second,y=G[i][o].first+x;
            while(y<=2ll*f[i+1])
                G[i+1].push_back(P(x,y)),y+=x;
        }
    }
    while(q--)
    {
        x=read(),y=read();if(x<y)swap(x,y);
        if(y==1){printf("1 %lld\n",x%mo);continue;}
        b=upper_bound(f+1,f+1+89,y)-f;
        if(x>=f[b])ans=b,printf("%d ",b);
        else ans=b-1,printf("%d ",b-1);
        if(ans==1){printf("%lld\n",1ll*x%mo*y%mo);continue;}
        sz=G[ans-1].size();res=0;
        for(o=0;o<sz;++o)
        {
            xx=G[ans-1][o].first;
            yy=G[ans-1][o].second;
            if(yy<=x)Add(res,1ll*(y-xx)/yy%mo);
            if(yy<=y)Add(res,1ll*(x-xx)/yy%mo);
        }
        printf("%d\n",res);
    }
	return 0;
}

### Prompt

Create a solution in cpp for 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;
#define reg register
#define ll long long
#define mo 1000000007
#define P make_pair
inline ll read()
{
	ll x=0,w=0;char ch=getchar();
	while(!isdigit(ch))w|=ch=='-',ch=getchar();
	while(isdigit(ch))x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
	return w?-x:x;
}
inline void Max(int &x,int y){x=x>y?x:y;}
int q,ans,res;
ll f[100],x,y;
vector< pair<ll,ll> >G[100];
inline void Add(int &x,int y){x+=y;if(x>=mo)x-=mo;}
int main()
{
    q=read();reg int i,o,sz,b;
    f[0]=1;f[1]=2;reg ll xx,yy;
    for(i=2;i<=89;++i)f[i]=f[i-1]+f[i-2];
    G[1].push_back(P(1,2));G[1].push_back(P(1,3));G[1].push_back(P(1,4));
    for(i=1;i<87;++i)
    {
        sz=G[i].size();
        for(o=0;o<sz;++o)
        {
            x=G[i][o].second,y=G[i][o].first+x;
            while(y<=2ll*f[i+1])
                G[i+1].push_back(P(x,y)),y+=x;
        }
    }
    while(q--)
    {
        x=read(),y=read();if(x<y)swap(x,y);
        if(y==1){printf("1 %lld\n",x%mo);continue;}
        b=upper_bound(f+1,f+1+89,y)-f;
        if(x>=f[b])ans=b,printf("%d ",b);
        else ans=b-1,printf("%d ",b-1);
        if(ans==1){printf("%lld\n",1ll*x%mo*y%mo);continue;}
        sz=G[ans-1].size();res=0;
        for(o=0;o<sz;++o)
        {
            xx=G[ans-1][o].first;
            yy=G[ans-1][o].second;
            if(yy<=x)Add(res,1ll*(y-xx)/yy%mo);
            if(yy<=y)Add(res,1ll*(x-xx)/yy%mo);
        }
        printf("%d\n",res);
    }
	return 0;
}

```