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;
#define ll long long
#define PA pair<ll,ll>
const int mod = 1000000007;
vector<PA >a[110];
ll f[110];
int Q;
void init()
{
	a[1].push_back(PA(1,1));
	f[0]=f[1]=1;
	for(int i=2;i<=90;i++)
		f[i]=f[i-1]+f[i-2];
	for(int i=2;i<90;i++)
	{
		a[i].push_back(PA(f[i-1],f[i]));
		a[i].push_back(PA(f[i-1],f[i+1]));
		for(int j=1;j<a[i-1].size();j++)
		{
			PA t=a[i-1][j];
			a[i].push_back(PA(t.second,t.first+t.second));
		}
	}
}
int main()
{
	//freopen("tt.in","r",stdin);
	init();
	scanf("%d",&Q);
	while(Q--)
	{
		ll x,y,ans=0;
		scanf("%lld%lld",&x,&y);
		if(x>y)swap(x,y);
		if(x==1)
		{
			printf("1 %lld\n",y%mod);
			continue;
		}
		if(x==2&&y==2)
		{
			puts("1 4");
			continue;
		}
		int k=upper_bound(f+1,f+90+1,x)-f;
		k=min(k,(int)(upper_bound(f+1,f+90+1,y)-f-1));
		
		for(int i=0;i<a[k-1].size();i++)
		{
			PA t=a[k-1][i];
			if(t.second<=x)
			{
				ans+=(y-t.first)/t.second;
				ans+=(x-t.first)/t.second;
			}
		}
		printf("%d %lld\n",k-1,ans%mod);
	}
	return 0;
}