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>
#define LL long long
#define PAIR pair<LL,LL>
using namespace std;
const int mo=1e9+7;
int n,p,q; LL ans,f[95];
vector<PAIR> V[95];
void Query()
{
	int i; LL x,y; p=1;
	scanf("%lld %lld",&x,&y);
	if(x>y) swap(x,y);
	while(f[p+1]<=x&&f[p+2]<=y) p++;
	printf("%d ",p);
	if(p==1){
		if(x==1){ printf("%d\n",(int)(y%mo));}
		else printf("4\n");
		return ;
	  }
	p--,ans=0;
	for(i=0;i<V[p].size();i++){
		if(V[p][i].second<=x) (ans+=(y-V[p][i].first)/V[p][i].second)%=mo;
		if(V[p][i].second<=y) (ans+=(x-V[p][i].first)/V[p][i].second)%=mo;
	  }
	printf("%d\n",ans);
}
void prepare()
{
	int i,j; LL k;
	V[1].push_back(PAIR(1,2));
	V[1].push_back(PAIR(1,3));
	for(i=2;i<=n;i++)
		for(j=0;j<V[i-1].size();j++)
			for(k=V[i-1][j].first+V[i-1][j].second;k<=f[i+2];k+=V[i-1][j].second)
				V[i].push_back(PAIR(V[i-1][j].second,k));
}
int main()
{
	scanf("%d",&q);
	f[0]=f[1]=1,n=1;
	while(f[n]+f[n-1]<=(1e18))
		f[n+1]=f[n]+f[n-1],++n;
	f[n+1]=f[n]+f[n-1];
	f[n+2]=f[n+1]+f[n];
	prepare();
	while(q--) Query();
	return 0;
}