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<cstdio>
#include<iostream>
using namespace std;
const long long maxv=1e18;
const long long md=1e9+7;
struct node{
	long long x,y;
}b[105][210];
int num[210];
long long f[105];
int ttp;
int m;
int main(){
//	freopen("F.in","r",stdin);
	f[0]=1,f[1]=2;
	for(int i=2;i<=100;i++)
	{
		f[i]=f[i-1]+f[i-2];
		if(f[i]>2*maxv){
			ttp=i;
			break;
		}
	}
	b[1][1]=(node){1,2};
	b[1][2]=(node){1,3};
	num[1]=2;
	long long x,y;
	for(int i=2;i<ttp;i++){
		for(int j=1;j<=num[i-1];j++)
		{
			x=b[i-1][j].x,y=b[i-1][j].y;
			while(x+y<=f[i+1]){
				x=x+y;
				b[i][++num[i]]=(node){y,x};
			}
		}
	}
	
	int M;
	scanf("%d",&M);
	long long s,t;
	int n;
	long long ans;
	for(m=1;m<=M;m++){
		ans=0;
		scanf("%lld%lld",&s,&t);
		if(s==1&&t==1){
			printf("1 1\n");
			continue;
		}
		if(s>t) swap(s,t);
		for(int i=1;i<=ttp;i++){
			if(f[i-1]<=s&&f[i]<=t);
			else{
				n=i-1;
				break;
			}
		}
		if(n==1){
			if(s>1&&t>1){
				printf("1 4\n");
			}
			else
				printf("1 %lld\n",(s+t-1)%md);
			continue;
		}
		for(int i=1;i<=num[n-1];i++){
			x=b[n-1][i].x;
			y=b[n-1][i].y;
			if(x<=s&&y<=s)
			{
				ans=(ans+(t-x)/y)%md;
				ans=(ans+(s-x)/y)%md;
			}
		}
		printf("%d %lld\n",n,ans);
	}
	return 0;
}