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: Runtime Error)

#include<stdio.h>
#include<algorithm>
using namespace std;
const int mod=1000000007;
int ad(int a,int b){return(a+b)%mod;}
int de(int a,int b){return(a-b)%mod;}
struct peo{
	int x,v;
}p[200010];
struct seg{
	int l,r;
}s[200010];
bool operator<(peo a,peo b){return a.x<b.x;}
bool operator<(seg a,seg b){return(a.r==b.r)?a.l<b.l:a.r<b.r;}
int fast[200010],slow[200010],f[200010],sum[200010];
int main(){
	int n,i,l,r,mid,ans,t;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d%d",&p[i].x,&p[i].v);
	sort(p+1,p+n+1);
	fast[1]=p[1].v;
	for(i=2;i<=n;i++)fast[i]=max(fast[i-1],p[i].v);
	slow[n]=p[n].v;
	for(i=n-1;i>0;i--)slow[i]=min(slow[i+1],p[i].v);
	for(i=1;i<=n;i++){
		ans=i;
		l=1;
		r=i-1;
		while(l<=r){
			mid=(l+r)>>1;
			if(fast[mid]>p[i].v){
				ans=mid;
				r=mid-1;
			}else
				l=mid+1;
		}
		s[i].l=ans;
		ans=i;
		l=i+1;
		r=n;
		while(l<=r){
			mid=(l+r)>>1;
			if(slow[mid]<p[i].v){
				ans=mid;
				l=mid+1;
			}else
				r=mid-1;
		}
		s[i].r=ans;
	}
	sort(s+1,s+n+1);
	t=0;
	f[0]=sum[0]=1;
	for(i=1;i<=n;i++){
		while(s[t+1].r<i)t++;
		f[i]=de(sum[i-1],sum[s[t].l-1]);
		sum[i]=ad(sum[i-1],f[i]);
	}
	printf("%d",ad(de(sum[n],sum[s[n].l-1]),mod));
}