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>
const int mo=1e9+7;
using namespace std;
typedef long long ll;
const int N=92;
ll f[N],x,y;
vector<pair<ll,ll>>v[N];
int Q,i,ans;
int main(){
	for(i=2,f[1]=1;i<N;++i)f[i]=f[i-1]+f[i-2];
	v[2]={make_pair(1,2)};
	for(i=3;i<N;++i){
		v[i]=v[i-1];
		for(auto&u:v[i])swap(u.first,u.second),u.second+=u.first;
		v[i].push_back(make_pair(f[i],f[i]*2+f[i-1]));
	}
	x=3;y=4;for(i=3;i<N;++i)v[i].push_back(make_pair(x,y)),swap(x,y),y+=x;
	for(scanf("%d",&Q);Q--;){
		scanf("%lld%lld",&x,&y);if(x>y)swap(x,y);
		if(x==1){printf("1 %lld\n",y%mo);continue;}
		if(x==2 && y==2){puts("1 4");continue;}
		for(i=3;f[i]<=x && f[i+1]<=y;++i);--i;ans=0;
		for(auto u:v[i])if(x>=u.first && y>=u.second && u.second<u.first*2)ans=(ans+(y-u.second)/u.first+1+
			(x>=u.second && y>=u.first))%mo;
		printf("%d %d\n",i-1,ans);
	}
	return 0;
}