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<iostream>
#include<cstring>
#include<cassert>
#include<cmath>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<time.h>
#include<algorithm>
using namespace std;
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(ll i=pos[x];~i;i=e[i].next)
#define ll long long
#define db double
const ll N=107;
const ll INF=1e18+7;
const ll mod=1e9+7;
ll Q,cnt,x,y,mx,Ans;
ll f[N];
bool check(ll a,ll b,ll st){
	ll X=f[st+1]*a+f[st]*b,Y=f[st]*a+f[st-1]*b;
	return (X>x)|(Y>y);
}
void dfs(ll a,ll b,ll st){
	if(check(a,b,mx-st))return;
	if(st==mx-1){
		if(x>=a)Ans=((y-b)/a%mod+Ans)%mod;
		if(y>=a)Ans=((x-b)/a%mod+Ans)%mod;
		return;
	}
	dfs(b+a,a,st+1);
	if(st==mx-2)dfs(b+a+a,a,st+1);
	else{
		ll p1=b+a+a,p2=a,p=mx-1-(st+1);
		dfs(f[p+1]*p1+f[p]*p2,f[p]*p1+f[p-1]*p2,mx-1);
	}
}
int main(){
	f[0]=0;
	f[1]=1;
	for(ll i=2;;i++){
		f[i]=f[i-1]+f[i-2];
		if(f[i]>INF){
			cnt=i;
			break;
		}
	}
	scanf("%lld",&Q);
	rep(gg,Q){
		scanf("%lld%lld",&x,&y);
		if(x==1&&y==1){
			puts("1 1");
			continue;
		}
		if(x<y)swap(x,y);
		mx=0;
		rep(i,cnt){
			if(f[i+1]>x||f[i]>y){
				mx=i-2;
				break;
			}
		}
		if(mx==1){
			printf("%lld %lld\n",mx,(x%mod)*(y%mod)%mod);
			continue;
		}
		Ans=0;
		dfs(1,1,0);
		printf("%lld %lld\n",mx,Ans);
	}
	return 0;
}