Problem:
Score : 1700 points

Problem Statement

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.

Sample Input 1

3
4 4
6 10
12 11

Sample Output 1

2 4
4 1
4 7

In the first query, (2,3), (3,2), (3,4) and (4,3) have a Euclidean step count of 2. No pairs have a step count of more than 2.

In the second query, (5,8) has a Euclidean step count of 4.

In the third query, (5,8), (8,5), (7,11), (8,11), (11,7), (11,8), (12,7) have a Euclidean step count of 4.

Sample Input 2

10
1 1
2 2
5 1000000000000000000
7 3
1 334334334334334334
23847657 23458792534
111111111 111111111
7 7
4 19
9 10

Sample Output 2

1 1
1 4
4 600000013
3 1
1 993994017
35 37447
38 2
3 6
3 9
4 2

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;

typedef long long LL;
typedef double db;
typedef unsigned long long ull;
const int mo =1e+9+7;

ull f[110];
LL n,m;

LL calc(LL x,LL y){
	LL cnt=0;
	if (x<=n&&y<=m)cnt=((m-y)/x+1)%mo;
	if (x<=m&&y<=n)cnt=(cnt+(n-y)/x+1)%mo;
	return cnt;
}

int main(){
	f[0]=f[1]=1;
	fo(i,2,90)f[i]=f[i-1]+f[i-2];
	int T;
	scanf("%d",&T);
	for(;T;T--){
		scanf("%lld%lld",&n,&m);
		if (n>m)swap(n,m);
		if (n==1&&m==1){printf("1 1\n");continue;}
		int k=0;
		fo(i,1,90)
		if (n>=f[i]&&m>=f[i+1])k=i;else break;
		if (k==1){
			if (n==2)printf("1 4\n");
			else printf("1 %lld\n",m%mo);
			continue;
		}
		printf("%d ",k);
		LL ans=0;
		ans=calc(f[k],f[k+1]);
		fo(i,1,k-1)ans=(ans+calc(f[k]+f[i]*f[k-1-i],f[k+1]+f[i]*f[k-i]))%mo;
		fo(i,1,k-1)
			fo(j,i,k-1)ans=(ans+calc(f[k]+f[i]*f[k-1-i]+f[j]*f[k-1-j],f[k+1]+f[i]*f[k-i]+f[j]*f[k-j]))%mo;
		printf("%lld\n",ans);
	}
	return 0;
}