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 <iostream>
#include <cstdio>
#include <cmath>
#include <map>
#include <vector>
#include <algorithm>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(itn i=a;i>=b;i--)
using namespace std;
const int Mod=1e9+7;
typedef long long ll;
ll ans,Ans,fib[105]={1,1},x,y;
vector<pair<ll,ll> >p[105];
int main(){
	rep(i,2,90) fib[i]=fib[i-1]+fib[i-2];
	//rep(i,1,100) if(fib[i]>1e18) printf("Here %d\n",i);
	p[1].push_back(make_pair(1,2));
	p[1].push_back(make_pair(1,3));
	p[1].push_back(make_pair(1,4));
	rep(i,1,90){
		for(int j=0;j<p[i].size();j++){
			ll u=p[i][j].second,v=p[i][j].first+u;
			for(;v<=fib[i+3]+fib[i];v+=u){
				p[i+1].push_back(make_pair(u,v));
			}
		}
	}
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%lld%lld",&x,&y);
		if(x>y) swap(x,y);
		for(ans=1,Ans=0;fib[ans+1]<=x&&fib[ans+2]<=y;ans++); printf("%d ",ans);
		if(ans==1){
			printf("%lld\n",x*y%Mod);
			continue;
		}
		for(int j=0;j<p[ans-1].size();j++){
			ll u=p[ans-1][j].first,v=p[ans-1][j].second;
			if(v<=x) Ans=(Ans+(y-u)/v%Mod)%Mod;
			if(v<=y) Ans=(Ans+(x-u)/v%Mod)%Mod;
		}
		printf("%lld\n",Ans);
	}
	return 0;
}