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<bits/stdc++.h>
#define ll long long
#define pi acos(-1)
#define pa pair<ll,ll>
#define mp make_pair
#define xx first
#define yy second
using namespace std;
const int N=105,mod=1e9+7;
const ll inf=1e18;
inline ll rd(){
	ll x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=0;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
	return f?x:-x;
}ll fib[N];vector<pa>a[N];
vector<pa>::iterator it;
ll x,y,n,num,m;int tot,tes,ans;
int main(){
	tes=rd();fib[0]=fib[1]=1;tot=1;
	while(1){
		fib[tot+1]=fib[tot]+fib[tot-1];
		if(fib[tot+1]>inf) break;
		tot++;
	}
	a[1].push_back(mp(1,2));a[1].push_back(mp(1,3));
	a[1].push_back(mp(1,4));
	for(int i=1;i<=tot-3;i++){
		it=a[i].begin();
		while(it!=a[i].end()){
			x=(*it).xx,y=(*it).yy;
			x+=y;
			while(x<=fib[i+3]+fib[i]){
				a[i+1].push_back(mp(y,x));
				x+=y;
			}
			it++;
		}
	}
	while(tes--){
		n=rd(),m=rd();
		if(n>m) swap(n,m);ans=1;
		while(fib[ans+1]<=n&&fib[ans+2]<=m) ans++;
		printf("%d ",ans);
		if(ans==1){
			printf("%lld\n",n*m%mod);
			continue;
		}
		else{
			num=0;it=a[ans-1].begin();
			while(it!=a[ans-1].end()){
				x=(*it).xx;y=(*it).yy;
				if(y<=n){
					num=(num+(m-x)/y)%mod;
				}
				if(y<=m){
					num=(num+(n-x)/y)%mod;
				}
				it++;
			}
			printf("%d\n",num);
		}
	}
	return 0;
}