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<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
#define inf 1e18
using namespace std;

const int maxn = 310;
const ll Mod = 1e9+7;

struct node
{
	ll x,y;
	node(){}
	node(const ll _x,const ll _y){x=_x;y=_y;}
}a[maxn][maxn]; int kn[maxn];
ll f[maxn];

void G(int k) // gernerate k+1
{
	for(int i=1;i<=kn[k];i++)
	{
		ll y=a[k][i].x,x=a[k][i].y;
		for(int j=1;;j++)
		{
			y+=x; if(y>f[k+3]) break;
			a[k+1][++kn[k+1]]=node(x,y);
		}
	}
}
ll solve(ll n,ll m,int k)
{
	ll re=0;
	if(k<=0)
	{
		for(int i=1;i<=n;i++)
		{
			ll k1=n/i,k2=m/i;
			(re+=k1*2ll%Mod-1+k2-k1)%=Mod;
		}
	}
	else
	{
		for(int i=1;i<=kn[k];i++)
		{
			ll y=a[k][i].x,x=a[k][i].y;
			// x px+y
			if(x>n) continue;
			ll k1=(n-y)/x,k2=(m-y)/x;
			(re+=k1*2ll%Mod+k2-k1)%=Mod;
		}
	}
	return re;
}

int main()
{
	f[0]=f[1]=1ll; for(int i=2;f[i-2]<=inf;i++) f[i]=f[i-1]+f[i-2];
	a[1][1]=node(1,2); a[1][2]=node(1,3); kn[1]=2;
	for(int i=2;f[i]<=inf;i++) 
		G(i-1);
	
	int t; scanf("%d",&t);
	while(t--)
	{
		ll x,y; scanf("%lld%lld",&x,&y); if(x>y) swap(x,y);
		int k;	for(k=1;f[k+1]<=y&&f[k]<=x;k++);k--;
		k--; if(k<0) k=0;
		ll ans=solve(x,y,k);
		printf("%d %lld\n",k+1,(ans+Mod)%Mod);
	}
	
	return 0;
}