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<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
LL gi() {
	LL w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
LL fib[100];
vector<pair<LL,LL>>v[100];
#define x first
#define y second
int main()
{
	int ans,i,Q=gi();LL x,y;
	
	fib[0]=fib[1]=1;
	for (i=2;(fib[i]=fib[i-1]+fib[i-2])<=1e18;i++);

	v[1].push_back(make_pair(1,2));
	v[1].push_back(make_pair(1,3));

	for (i=1;i<=90;i++)
		for (auto t:v[i-1])
			for (x=t.y,y=t.x+t.y;y<=fib[i+2]+fib[i-1];y+=t.y)
				v[i].push_back(make_pair(x,y));

	const int mod=1e9+7;
	
	while (Q--) {
		x=gi(),y=gi();
		if (x>y) swap(x,y);
		if (x==1) printf("1 %d\n",y%mod);
		else if (y==2) puts("1 4");
		else {
			for (i=0;fib[i]<=x&&fib[i+1]<=y;i++);i--;
			printf("%d ",i);
			ans=0;
			for (auto t:v[i-1]) {
				if (t.y<=x)
					ans=(ans+(y-t.x)/t.y)%mod;
				if (t.y<=y)
					ans=(ans+(x-t.x)/t.y)%mod;
			}
			printf("%d\n",ans);
		}
	}
	
	return 0;
}