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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<stdlib.h>
#include<cassert>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=500000004;
const double EPS=1e-10;
const double PI=acos(-1.0);
int ABS(int a){return max(a,-a);}
long long ABS(long long a){return max(a,-a);}
long long fib[100];
vector<pair<long long,long long > > v[100];
void dfs(int a,long long b,long long c,int d){
	if(b&&c/b==1){
		v[a].push_back(make_pair(b,c));
		v[a].push_back(make_pair(c,b));
	}
	if(d==0){
		long long tb=c;
		long long tc=b+c*2;
		if(tb<=fib[a+3]&&tc<=fib[a+3])dfs(a+1,tb,tc,1);
	}
	long long tb=c;
	long long tc=b+c;
	if(tb<=fib[a+3]&&tc<=fib[a+3])dfs(a+1,tb,tc,d);
}
int main(){
	int a;scanf("%d",&a);
	fib[1]=1;
	fib[2]=2;
	for(int i=3;i<100;i++){
		fib[i]=min(inf,fib[i-2]+fib[i-1]);
	}
	dfs(0,1,1,0);
	for(int i=1;i<=4;i++){
//		for(int j=0;j<v[i].size();j++)printf("%d: %lld %lld\n",i,v[i][j].first,v[i][j].second);
	}
	while(a--){
		long long x,y;scanf("%lld%lld",&x,&y);
		if(x>y)swap(x,y);
		int K=1;
		for(int i=98;i>=1;i--){
			if(fib[i]<=x&&fib[i+1]<=y){
				K=i;
				break;
			}
		}
		printf("%d ",K);
		if(K==1){
			printf("%lld\n",(x%mod)*(y%mod)%mod);
			continue;
		}else{
			long long ret=0;
			for(int i=0;i<v[K].size();i++){
				if(v[K][i].first>x||v[K][i].second>y)continue;
				ret=(ret+(y-v[K][i].second)/v[K][i].first+1)%mod;
			}
			printf("%lld\n",ret);
		}
	}
}