//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=100+7,W=90;
const ll mod=1e9+7;
ll Td,n,m,ans,sum,f[maxn];

char cc;ll ff;
template<typename T>void read(T& aa) {
	aa=0;ff=1; cc=getchar();
	while(cc!='-'&&(cc<'0'||cc>'9')) cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

struct Node{
	ll x,y;
	Node(){}
	Node(ll x,ll y):x(x),y(y){}
}o;
vector<Node> G[maxn];

int main() {
	ll x,y,s; f[0]=f[1]=1;
	For(i,2,W) f[i]=f[i-1]+f[i-2];
	G[1].push_back(Node(1,2)); G[1].push_back(Node(1,3));
	G[1].push_back(Node(1,4));
	For(i,1,W-3) {
		s=G[i].size();
		For(j,0,s-1) {
			o=G[i][j]; x=o.y; y=o.x+x;
			while(y<=f[i+3]+f[i]) {
				G[i+1].push_back(Node(x,y));
				y+=x;
			}
		}
	}
	
	read(Td);
	while(Td--) {
		read(n); read(m);
		if(n>m) swap(n,m);
		for(ans=1;f[ans+1]<=n&&f[ans+2]<=m;++ans) ;
		printf("%lld ",ans);
		if(ans==1) {
			printf("%lld\n",n*m%mod);
			continue;
		}
		s=G[ans-1].size(); sum=0;
		For(i,0,s-1) {
			o=G[ans-1][i]; x=o.x; y=o.y;
			if(y<=n) sum+=(m-x)/y%mod;
			if(y<=m) sum+=(n-x)/y%mod;
			sum%=mod;
		}
		printf("%lld\n",sum);
	}
	return 0;
}

### Prompt

Generate a CPP solution to 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
//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
#define ll long long
#define db double
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
const int maxn=100+7,W=90;
const ll mod=1e9+7;
ll Td,n,m,ans,sum,f[maxn];

char cc;ll ff;
template<typename T>void read(T& aa) {
	aa=0;ff=1; cc=getchar();
	while(cc!='-'&&(cc<'0'||cc>'9')) cc=getchar();
	if(cc=='-') ff=-1,cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	aa*=ff;
}

struct Node{
	ll x,y;
	Node(){}
	Node(ll x,ll y):x(x),y(y){}
}o;
vector<Node> G[maxn];

int main() {
	ll x,y,s; f[0]=f[1]=1;
	For(i,2,W) f[i]=f[i-1]+f[i-2];
	G[1].push_back(Node(1,2)); G[1].push_back(Node(1,3));
	G[1].push_back(Node(1,4));
	For(i,1,W-3) {
		s=G[i].size();
		For(j,0,s-1) {
			o=G[i][j]; x=o.y; y=o.x+x;
			while(y<=f[i+3]+f[i]) {
				G[i+1].push_back(Node(x,y));
				y+=x;
			}
		}
	}
	
	read(Td);
	while(Td--) {
		read(n); read(m);
		if(n>m) swap(n,m);
		for(ans=1;f[ans+1]<=n&&f[ans+2]<=m;++ans) ;
		printf("%lld ",ans);
		if(ans==1) {
			printf("%lld\n",n*m%mod);
			continue;
		}
		s=G[ans-1].size(); sum=0;
		For(i,0,s-1) {
			o=G[ans-1][i]; x=o.x; y=o.y;
			if(y<=n) sum+=(m-x)/y%mod;
			if(y<=m) sum+=(n-x)/y%mod;
			sum%=mod;
		}
		printf("%lld\n",sum);
	}
	return 0;
}

```