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>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int Q;
ll F[91];
ll X,Y;

vector<pair<ll,ll>> cand[91];
ll mo=1000000007;

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	F[0]=F[1]=1;
	for(i=2;i<=89;i++) {
		F[i]=F[i-1]+F[i-2];
		for(j=2;j<=i+1;j++) {
			ll a=1,b=1;
			for(x=2;x<=i;x++) {
				ll c=b+a;
				if(x==j) c+=a;
				b=a;
				a=c;
			}
			cand[i].push_back({a,a+b});
			cand[i].push_back({a+b,a});
		}
	}
	
	
	cin>>Q;
	while(Q--) {
		cin>>X>>Y;
		
		if(X>Y) swap(X,Y);
		if(X==1 || (X==2&&Y==2)) {
			cout<<1<<" "<<(X*Y)%mo<<endl;
			continue;
		}
		
		int k=0;
		while(F[k+1]<=X && F[k+2]<=Y) k++;
		ll ret=0;
		FORR(r,cand[k]) if(X>=r.first && Y>=r.second) {
			(ret += 1+(Y-r.second)/r.first)%=mo;
		}
		cout<<k<<" "<<ret<<endl;
		
	}
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}