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 mp  make_pair
#define pb  push_back
using namespace std;

typedef long long  LL;
typedef pair <LL,LL>  PLL;
const int mo=int(1e9+7);
LL x,y,ans;
vector <PLL> a[100];

void prepare()
{
	a[0].pb(mp(1,1));
	for (int i=1; i<100; i++)
		{
			for (int j=0; j<a[i-1].size(); j++)
				a[i].pb(mp(a[i-1][j].second,a[i-1][j].first+a[i-1][j].second));
			a[i].pb(mp(a[i-1][0].second,a[i-1][0].first+2*a[i-1][0].second));
		}
}

void work()
{
	scanf("%lld %lld",&x,&y),ans=0;
	if (x>y)  swap(x,y);
	for (int i=1; i<100; i++)
		if ((x<a[i+1][0].first)||(y<a[i+1][0].second))
			{
				if (i==1)  {printf("%d %lld\n",i,x*y%mo);  return;}
				for (int j=0; j<a[i-1].size(); j++)
					{
						LL p=a[i-1][j].first,q=a[i-1][j].second;
						if ((q<=x)&&(p+q<=y))  ans=(ans+(y-p)/q)%mo;
						if ((q<=y)&&(p+q<=x))  ans=(ans+(x-p)/q)%mo;
					}
				printf("%d %lld\n",i,ans);
				return;
			}
}

int main()
{
	int t;
	scanf("%d",&t),prepare();
	while (t--)  work();
	return 0;
}