#include <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<P,int> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1000000000000000000;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

int q,n;
ll t,x,y;
vl a,b;

ll f(int I,ll A,ll B){
	if(I==1) b[I]=2;
	else b[I]=b[I-1]+b[I-2];
	if(I==t){
		ll res=0;
		if(b[I]<=x) (res+=(x-b[I])/b[I-1]+1)%=mod;
		if(b[I]<=y) (res+=(y-b[I])/b[I-1]+1)%=mod;
		return res;
	}
	ll res=f(I+1,A,B);
	while(1){
		b[I]+=b[I-1];
		A-=b[I-1]*(t-I-2<0?1:a[t-I-2]);
		B-=b[I-1]*(t-I-1<0?1:a[t-I-1]);
		if(A<0||B<0) break;
		(res+=f(I+1,A,B))%=mod;
	}
	return res;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>q;
	a.push_back(1);
	a.push_back(2);
	while(a.back()<=INF) a.push_back(a.back()+a[a.size()-2]);
	n=a.size();
	b=a;
	for(int i=0;i<q;i++){
		cin>>x>>y;
		if(x==1&&y==1){
			cout<<1<<' '<<1<<endl;
			continue;
		}
		if(x>y) swap(x,y);
		for(int j=0;j<n;j++) if(a[j]>x){
			t=j;
			if(a[j]>y) t--;
			break;
		}
		cout<<t<<' '<<f(1,x-a[t-1],y-a[t])+(t==1?x:0)<<endl;
	}
}
### 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>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<P,int> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1000000000000000000;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

int q,n;
ll t,x,y;
vl a,b;

ll f(int I,ll A,ll B){
	if(I==1) b[I]=2;
	else b[I]=b[I-1]+b[I-2];
	if(I==t){
		ll res=0;
		if(b[I]<=x) (res+=(x-b[I])/b[I-1]+1)%=mod;
		if(b[I]<=y) (res+=(y-b[I])/b[I-1]+1)%=mod;
		return res;
	}
	ll res=f(I+1,A,B);
	while(1){
		b[I]+=b[I-1];
		A-=b[I-1]*(t-I-2<0?1:a[t-I-2]);
		B-=b[I-1]*(t-I-1<0?1:a[t-I-1]);
		if(A<0||B<0) break;
		(res+=f(I+1,A,B))%=mod;
	}
	return res;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>q;
	a.push_back(1);
	a.push_back(2);
	while(a.back()<=INF) a.push_back(a.back()+a[a.size()-2]);
	n=a.size();
	b=a;
	for(int i=0;i<q;i++){
		cin>>x>>y;
		if(x==1&&y==1){
			cout<<1<<' '<<1<<endl;
			continue;
		}
		if(x>y) swap(x,y);
		for(int j=0;j<n;j++) if(a[j]>x){
			t=j;
			if(a[j]>y) t--;
			break;
		}
		cout<<t<<' '<<f(1,x-a[t-1],y-a[t])+(t==1?x:0)<<endl;
	}
}
```