#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define MAXN 110
#define LL long long
#define P 1000000007
using namespace std;

LL nfib,fib[MAXN];
vector< pair<LL,LL> > S[MAXN];
LL n1,n2;
int nl,cnt,dep;

void init(){
	nfib=90;
	fib[0]=1;
	fib[1]=1;
	for(int i=2;i<=nfib;i++) fib[i]=fib[i-1]+fib[i-2];
	S[1].push_back(make_pair(1,2)); S[1].push_back(make_pair(2,4));
	for(int i=2;i<=nfib;i++){
		for(int j=0;j<S[i-1].size();j++){
			LL x1=S[i-1][j].first,x2=S[i-1][j].second;
			while(x2<fib[i+2]){
				S[i].push_back(make_pair(x2,x1+x2));
				x2+=x1;
			}
		}
	}
}

int gao1(LL X,LL Y){
	if(X==1 && Y==1) return 1;
	int now=1;
	while(fib[now]<=X && fib[now+1]<=Y) now++;
	return --now;
}

pair<LL,LL> trans(int step,LL x1,LL x2){
	LL k1=(step>=2?fib[step-2]:0);
	LL k2=(step>=1?fib[step-1]:0);
	LL k3=(step>=1?fib[step-1]:0);
	LL k4=fib[step];
	if(!step) k1=1;
	return make_pair(k1*x1+k3*x2,k2*x1+k4*x2);
}

LL gao(LL X,LL Y){
	if(fib[nl]>X || fib[nl+1]>Y) return 0;
	LL res=0;
	for(int i=0;i<S[nl].size();i++){
		LL x1=S[nl][i].first,x2=S[nl][i].second;
		if(x1<=X && x2<=Y) res+=(Y-x2)/x1+1;
	}
	return res;
}

void query(LL X,LL Y){
	if(X>Y) swap(X,Y);
	nl=gao1(X,Y);
	printf("%lld ",nl);
	if(nl==1){
		printf("%lld\n",X*Y%P);
		return;
	}
	LL ans2=(gao(X,X)+gao(X,Y))%P;
	printf("%lld\n",ans2);
}

int main(){
#ifdef DEBUG
	freopen("F.in","r",stdin);
#endif
	init();
	int m;
	scanf("%d",&m);
	while(m--){
		LL X,Y;
		scanf("%lld%lld",&X,&Y);
		query(X,Y);
	}
}

### 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 <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define MAXN 110
#define LL long long
#define P 1000000007
using namespace std;

LL nfib,fib[MAXN];
vector< pair<LL,LL> > S[MAXN];
LL n1,n2;
int nl,cnt,dep;

void init(){
	nfib=90;
	fib[0]=1;
	fib[1]=1;
	for(int i=2;i<=nfib;i++) fib[i]=fib[i-1]+fib[i-2];
	S[1].push_back(make_pair(1,2)); S[1].push_back(make_pair(2,4));
	for(int i=2;i<=nfib;i++){
		for(int j=0;j<S[i-1].size();j++){
			LL x1=S[i-1][j].first,x2=S[i-1][j].second;
			while(x2<fib[i+2]){
				S[i].push_back(make_pair(x2,x1+x2));
				x2+=x1;
			}
		}
	}
}

int gao1(LL X,LL Y){
	if(X==1 && Y==1) return 1;
	int now=1;
	while(fib[now]<=X && fib[now+1]<=Y) now++;
	return --now;
}

pair<LL,LL> trans(int step,LL x1,LL x2){
	LL k1=(step>=2?fib[step-2]:0);
	LL k2=(step>=1?fib[step-1]:0);
	LL k3=(step>=1?fib[step-1]:0);
	LL k4=fib[step];
	if(!step) k1=1;
	return make_pair(k1*x1+k3*x2,k2*x1+k4*x2);
}

LL gao(LL X,LL Y){
	if(fib[nl]>X || fib[nl+1]>Y) return 0;
	LL res=0;
	for(int i=0;i<S[nl].size();i++){
		LL x1=S[nl][i].first,x2=S[nl][i].second;
		if(x1<=X && x2<=Y) res+=(Y-x2)/x1+1;
	}
	return res;
}

void query(LL X,LL Y){
	if(X>Y) swap(X,Y);
	nl=gao1(X,Y);
	printf("%lld ",nl);
	if(nl==1){
		printf("%lld\n",X*Y%P);
		return;
	}
	LL ans2=(gao(X,X)+gao(X,Y))%P;
	printf("%lld\n",ans2);
}

int main(){
#ifdef DEBUG
	freopen("F.in","r",stdin);
#endif
	init();
	int m;
	scanf("%d",&m);
	while(m--){
		LL X,Y;
		scanf("%lld%lld",&X,&Y);
		query(X,Y);
	}
}

```