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 <vector>
#include <utility>
#include <iostream>
using namespace std;
#define ref(i,x,y)for(int i=x;i<=y;++i)
typedef long long LL;
typedef pair<LL,LL> PII;
const int mod=1e9+7;
LL fib[101],x,y;
vector<PII> s[101];
int Q;
void dfs(int x,LL a,LL b){
	if(x>88)return;
	if(a<0||b<0)printf("error\n");
	if(a>=fib[x]&&b>=fib[x+1])return;
	s[x].push_back(make_pair(a,b));
	dfs(x+1,b,b+a);
	dfs(x+1,b,b*2+a);
}
int main(){
	fib[0]=fib[1]=1;
	ref(i,2,90)fib[i]=fib[i-1]+fib[i-2];
	dfs(1,1,1);
	scanf("%d",&Q);
	ref(i,1,Q){
		scanf("%lld%lld",&x,&y);
		if(x>y)swap(x,y);
		int ans=0;LL sum=0;
		ref(i,1,100)if(fib[i]>x||fib[i+1]>y){ans=i-1;break;}
		if(ans<=1){
			if(x==1)sum=y%mod;else if(x==2)sum=4;
			printf("1 %lld\n",sum);
			continue;
		}
		int sz=s[ans].size();
		ref(i,0,sz-1){
			PII p=s[ans][i];
			if(p.second<=x&&y>=p.first)(sum+=(y-p.first)/p.second)%=mod;
			if(p.second<=y&&x>=p.first)(sum+=(x-p.first)/p.second)%=mod;
		}
		printf("%d %lld\n",ans,sum);
	}
}