#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ui unsigned int
#define ull unsigned long long
#define db long double
#define pii pair<ll,ll>
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define vi vector<int>
#define vii vector<vi>
#define lb lower_bound
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,b,a) for(int i=(b);i>=(a);--i)
#define rep0(i,a,b) for(int i=(a);i<(b);++i)
#define fore(i,a) for(int i=0;i<a.size();++i)
#define gc() getchar()
#define ls x<<1,l,m
#define rs x<<1|1,m+1,r
inline ll rd()
{
	ll x=0;char c=gc();while(!isdigit(c))c=gc();
	while(isdigit(c))x=x*10+c-48,c=gc();return x;
}
const int N=1000005,B=90,P=1e9+7;
ll f[B+5];vector<pii>V[B+5];
int main()
{
	f[0]=f[1]=1;rep(i,2,B)f[i]=f[i-1]+f[i-2];
	V[1].pb(mp(1,2));V[1].pb(mp(1,3));V[1].pb(mp(1,4));
	rep(i,1,B-3)fore(j,V[i])
	{
		ll x=V[i][j].X,y=V[i][j].Y;y+=x;x=y-x;
		while(y<=f[i+3]+f[i-1])V[i+1].pb(mp(x,y)),y+=x;
	}
    int T=rd();
    while(T--)
    {
    	ll x=rd(),y=rd(),K=1;if(x>y)swap(x,y);
		while(f[K+1]<=x&&f[K+2]<=y)K++;printf("%d ",K);
		if(K==1){printf("%d\n",x*y%P);continue;}
		ll ans=0;
		fore(i,V[K-1]) 
		{
			ll a=V[K-1][i].X,b=V[K-1][i].Y;
			if(b<=x)ans=(ans+(y-a)/b)%P;
			if(b<=y)ans=(ans+(x-a)/b)%P;
		}
        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<bits/stdc++.h>
using namespace std;
#define ll long long
#define ui unsigned int
#define ull unsigned long long
#define db long double
#define pii pair<ll,ll>
#define X first
#define Y second
#define mp make_pair
#define pb push_back
#define vi vector<int>
#define vii vector<vi>
#define lb lower_bound
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,b,a) for(int i=(b);i>=(a);--i)
#define rep0(i,a,b) for(int i=(a);i<(b);++i)
#define fore(i,a) for(int i=0;i<a.size();++i)
#define gc() getchar()
#define ls x<<1,l,m
#define rs x<<1|1,m+1,r
inline ll rd()
{
	ll x=0;char c=gc();while(!isdigit(c))c=gc();
	while(isdigit(c))x=x*10+c-48,c=gc();return x;
}
const int N=1000005,B=90,P=1e9+7;
ll f[B+5];vector<pii>V[B+5];
int main()
{
	f[0]=f[1]=1;rep(i,2,B)f[i]=f[i-1]+f[i-2];
	V[1].pb(mp(1,2));V[1].pb(mp(1,3));V[1].pb(mp(1,4));
	rep(i,1,B-3)fore(j,V[i])
	{
		ll x=V[i][j].X,y=V[i][j].Y;y+=x;x=y-x;
		while(y<=f[i+3]+f[i-1])V[i+1].pb(mp(x,y)),y+=x;
	}
    int T=rd();
    while(T--)
    {
    	ll x=rd(),y=rd(),K=1;if(x>y)swap(x,y);
		while(f[K+1]<=x&&f[K+2]<=y)K++;printf("%d ",K);
		if(K==1){printf("%d\n",x*y%P);continue;}
		ll ans=0;
		fore(i,V[K-1]) 
		{
			ll a=V[K-1][i].X,b=V[K-1][i].Y;
			if(b<=x)ans=(ans+(y-a)/b)%P;
			if(b<=y)ans=(ans+(x-a)/b)%P;
		}
        printf("%lld\n",ans);
	}
	return 0;
}
```