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;

const int N=200,mod=1000000007;
typedef long long ll;
const ll inf=1e18;
inline void reduce(int&x){x+=x>>31&mod;}

vector<pair<ll,ll> >vec[N];
ll f[N];
int Q,cnt;
void Init(){
	f[0]=f[1]=1;cnt=1;
	for(;f[cnt]<=inf;++cnt)f[cnt+1]=f[cnt-1]+f[cnt];--cnt;
	for(int i=2;i<=f[3]+f[0];++i)vec[1].emplace_back(1,i);
	for(int i=2;i<=cnt;++i){
		for(auto j:vec[i-1]){
			auto t=make_pair(j.second,j.first+j.second);
			for(;t.second<=f[i-1]+f[i+2]&&t.second<=inf;t.second+=t.first)vec[i].push_back(t);
		}
	}
}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	Init();
	for(cin>>Q;Q --> 0;){
		ll x,y;cin>>x>>y;if(x>y)swap(x,y);
		int res=1,ans=0;
		for(;f[res]<=x&&f[res+1]<=y;++res);if(res>1)--res;
		if(res==1)ans=x*y%mod;
		else{
			for(auto i:vec[res-1]){
				if(i.second<=y)reduce(ans+=(x-i.first)/i.second%mod-mod);
				if(i.second<=x)reduce(ans+=(y-i.first)/i.second%mod-mod);
			}
		}
		cout<<res<<' '<<ans<<'\n';
	}
	return 0;
}