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 fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
inline ll read(){
	ll n=0,f=1;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if (c=='-') c=getchar(),f=-1;
	for(;c>='0'&&c<='9';c=getchar()) n=n*10+c-48;
	return n*f;
}
const int mo=1e9+7;
int Q,i,j,ans;
ll X,Y,inf=1e18+1,f[105];
struct ar{ll x,y;}now;
vector<ar>p[105];
int main(){
	f[0]=f[1]=1;
	fo(i,2,89) f[i]=f[i-1]+f[i-2];
	p[1].push_back((ar){1,2});
	p[1].push_back((ar){1,3});
	fo(i,1,86){
		for(j=0;j<p[i].size();j++) now=p[i][j],p[i+1].push_back((ar){now.y,now.y+now.x});
		now=p[i][0];
		p[i+1].push_back((ar){now.y,now.y*2+now.x});
	}
	Q=read();
	for(;Q--;){
		X=read(),Y=read();
		if (X>Y) swap(X,Y);
		ans=1;
		for(;f[ans+1]<=X&&f[ans+2]<=Y;ans++);
		printf("%d ",ans);
		if (ans==1){
			printf("%d\n",X*Y%mo);
			continue;
		}
		ans--;
		ll num=0;
		for(i=0;i<p[ans].size();i++){
			now=p[ans][i];
			if (now.y<=X) (num+=(Y-now.x)/now.y)%=mo;
			if (now.y<=Y) (num+=(X-now.x)/now.y)%=mo;
		}
		printf("%d\n",num);
	}
}