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>
typedef long long ll;
const int p=1e9+7;

void up(int &a,int b){(a+=b)>=p&&(a-=p);}

ll f[101][3];

int T,k;
ll x,y,ans;

void dfs(ll a,ll b,int t)
{
    if(t+1==k){ans+=((y-a)/b+(x-a)/b)%p;return;}

    for(a+=b;f[k-t][1]*b+f[k-t][2]*a<=x&&f[k-t+1][1]*b+f[k-t+1][2]*a<=y;a+=b)
	dfs(b,a,t+1);
}

int main()
{
    f[1][0]=f[1][1]=f[2][0]=f[2][2]=1;
    for(int i=3;i<=100;++i)
	for(int j=0;j<=2;++j)
	    f[i][j]=f[i-2][j]+f[i-1][j];
    
    scanf("%d",&T);
    for(;T--;)
    {
	scanf("%lld%lld",&x,&y);
	if(x>y)std::swap(x,y);

	if(x<2){printf("1 %lld\n",y%p);continue;}

	for(k=2;f[k][0]<=x&&f[k+1][0]<=y;++k);
	if((k-=2)<2){printf("1 %lld\n",x*y%p);continue;}

	ans=0;
	for(int i=2;f[k][1]+f[k][2]*i<=x&&f[k+1][1]+f[k+1][2]*i<=y;++i)
	    dfs(1,i,1);
	printf("%d %lld\n",k,ans%p);
    }
    
    return 0;
}