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 rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define per(i,n) for ((i)=n;(i)>=(1);(i)--)
using namespace std;
int q,s1,s2,mod=1e9+7;
long long a,b,f[205];
void dfs(long long x,long long y,int dep)
{
	if(dep+1==s1)
	{
		if(x<=a&&y<=b) s2=(s2+(a-x)/y)%mod;
		if(x<=b&&y<=a) s2=(s2+(b-x)/y)%mod;
		return;
	}
	while(x+y<=f[dep+3])
	{
		x+=y;
		dfs(y,x,dep+1);
	}
}
int main()
{
	int i;f[0]=f[1]=1;for(i=2;i<=200;i++) f[i]=f[i-1]+f[i-2];
	scanf("%d",&q);
	while(q--)
	{
		scanf("%lld%lld",&a,&b);if(a>b) swap(a,b);
		if(a==1){printf("1 %lld\n",b%mod);continue;} 
		if(a==2&&b==2){puts("1 4");continue;}
		s1=s2=0;while(f[s1+1]<=a&&f[s1+2]<=b) s1++;
		dfs(1,1,0);printf("%d %d\n",s1,s2);
	}
	return 0;
}