#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
int q;
const ll MOD=1e9+7;
#define S 91
ll fib[S];
typedef pair<ll,ll> pll;
#define pb push_back
#define fi first
#define se second
vector<pll> ex[S];
void sol()
{
	ll a,b;
	scanf("%lld%lld",&a,&b);
	if(a>b) swap(a,b);
	int ans=1;
	for(int i=2;i<S;i++)
	{
		if(fib[i]<=a&&fib[i+1]<=b) ans=i;
		else break;
	}
	printf("%d ",ans);
	ll tot=0;
	if(ans==1)
		tot=(a%MOD)*(b%MOD)%MOD; //small enough
	else
	{
		for(unsigned _=0;_<ex[ans-1].size();++_)
		{
			ll x=ex[ans-1][_].fi,y=ex[ans-1][_].se;
			//(y,x+ky)
			if(y<=a) tot+=(b-x)/y, tot%=MOD;
			if(y<=b) tot+=(a-x)/y, tot%=MOD;
		}
	}
	tot=(tot%MOD+MOD)%MOD;
	printf("%d\n",int(tot));
}
int calc(ll x,ll y)
{
	if(x<y) swap(x,y);
	if(x&&y) return calc(y,x%y)+1;
	return 0;
}
void gen(int x)
{
	vector<pll>&s=ex[x-1],&t=ex[x];
	for(unsigned _=0;_<s.size();++_)
	{
		pll g=s[_];
		if(g.fi>g.se) throw "!!!";
		ll tmp=g.se; swap(g.fi,g.se);
		for(g.se+=tmp;g.se<=fib[x+2];g.se+=tmp)
		{
			if(calc(g.fi,g.se)!=x) continue;
			t.pb(g);
		}
	}
	sort(t.begin(),t.end());
	t.erase(unique(t.begin(),t.end()),t.end());
}
int main()
{
	fib[0]=1; fib[1]=1;
	for(int i=2;i<S;i++) fib[i]=fib[i-1]+fib[i-2];
	ex[1].pb(pll(1,2)); ex[1].pb(pll(1,3));
	for(int i=2;i+2<S;i++)
		gen(i);
	scanf("%d",&q);
	while(q--) sol();
}
### Prompt

In cpp, your task is to solve 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 <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
int q;
const ll MOD=1e9+7;
#define S 91
ll fib[S];
typedef pair<ll,ll> pll;
#define pb push_back
#define fi first
#define se second
vector<pll> ex[S];
void sol()
{
	ll a,b;
	scanf("%lld%lld",&a,&b);
	if(a>b) swap(a,b);
	int ans=1;
	for(int i=2;i<S;i++)
	{
		if(fib[i]<=a&&fib[i+1]<=b) ans=i;
		else break;
	}
	printf("%d ",ans);
	ll tot=0;
	if(ans==1)
		tot=(a%MOD)*(b%MOD)%MOD; //small enough
	else
	{
		for(unsigned _=0;_<ex[ans-1].size();++_)
		{
			ll x=ex[ans-1][_].fi,y=ex[ans-1][_].se;
			//(y,x+ky)
			if(y<=a) tot+=(b-x)/y, tot%=MOD;
			if(y<=b) tot+=(a-x)/y, tot%=MOD;
		}
	}
	tot=(tot%MOD+MOD)%MOD;
	printf("%d\n",int(tot));
}
int calc(ll x,ll y)
{
	if(x<y) swap(x,y);
	if(x&&y) return calc(y,x%y)+1;
	return 0;
}
void gen(int x)
{
	vector<pll>&s=ex[x-1],&t=ex[x];
	for(unsigned _=0;_<s.size();++_)
	{
		pll g=s[_];
		if(g.fi>g.se) throw "!!!";
		ll tmp=g.se; swap(g.fi,g.se);
		for(g.se+=tmp;g.se<=fib[x+2];g.se+=tmp)
		{
			if(calc(g.fi,g.se)!=x) continue;
			t.pb(g);
		}
	}
	sort(t.begin(),t.end());
	t.erase(unique(t.begin(),t.end()),t.end());
}
int main()
{
	fib[0]=1; fib[1]=1;
	for(int i=2;i<S;i++) fib[i]=fib[i-1]+fib[i-2];
	ex[1].pb(pll(1,2)); ex[1].pb(pll(1,3));
	for(int i=2;i+2<S;i++)
		gen(i);
	scanf("%d",&q);
	while(q--) sol();
}
```