#include<bits/stdc++.h>
typedef long long ll;
#define mod 1000000007
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
ll fib[100];
std::vector<std::pair<ll,ll>>S[100];
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	fib[0]=fib[1]=1;for(int i=2;i<100;++i)fib[i]=fib[i-1]+fib[i-2];
	S[1].push_back({1,2});S[1].push_back({1,3});S[1].push_back({1,4});
	for(int i=1;i<99;++i){
		for(auto j:S[i]){
			ll x=j.second,y=j.first;
			y+=x;while(y<=fib[i+3]+fib[i])S[i+1].push_back({x,y}),y+=x;
		}
	}
	int Q=gi();
	while(Q--){
		ll X=gi(),Y=gi();if(X>Y)std::swap(X,Y);
		if(Y==1)puts("1 1");
		else{
			int p=1;
			while(X>=fib[p+1]&&Y>=fib[p+2])++p;
			if(p==1){printf("1 %d\n",X*Y%mod);continue;}
			int ans=0;
			for(auto pr:S[p-1]){
				ll x=pr.first,y=pr.second;
				if(y<=X)ans=(ans+(Y-x)/y)%mod;
				if(y<=Y)ans=(ans+(X-x)/y)%mod;
			}
			printf("%d %d\n",p,ans);
		}
	}
	return 0;
}

### Prompt

Create a solution in Cpp for 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<bits/stdc++.h>
typedef long long ll;
#define mod 1000000007
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
ll fib[100];
std::vector<std::pair<ll,ll>>S[100];
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	fib[0]=fib[1]=1;for(int i=2;i<100;++i)fib[i]=fib[i-1]+fib[i-2];
	S[1].push_back({1,2});S[1].push_back({1,3});S[1].push_back({1,4});
	for(int i=1;i<99;++i){
		for(auto j:S[i]){
			ll x=j.second,y=j.first;
			y+=x;while(y<=fib[i+3]+fib[i])S[i+1].push_back({x,y}),y+=x;
		}
	}
	int Q=gi();
	while(Q--){
		ll X=gi(),Y=gi();if(X>Y)std::swap(X,Y);
		if(Y==1)puts("1 1");
		else{
			int p=1;
			while(X>=fib[p+1]&&Y>=fib[p+2])++p;
			if(p==1){printf("1 %d\n",X*Y%mod);continue;}
			int ans=0;
			for(auto pr:S[p-1]){
				ll x=pr.first,y=pr.second;
				if(y<=X)ans=(ans+(Y-x)/y)%mod;
				if(y<=Y)ans=(ans+(X-x)/y)%mod;
			}
			printf("%d %d\n",p,ans);
		}
	}
	return 0;
}

```