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<bits/stdc++.h>
#define ll long long
using namespace std;

int n; ll ans,A,B,f[109];
bool ok(ll x,ll y,int now){
	now=n-now;
	return x*f[now-1]+y*f[now]<=A && x*f[now]+y*f[now+1]<=B;
}
void solve(ll x,ll y,int now){
	if (!ok(x,y,now)) return;
	if (now==n){
		ans++; return;
	}
	if (now+1==n && y<=A && x<=B){
		ans+=max(0ll,(B-x)/y); return;
	}
	solve(y,x+y,now+1);
	solve(y,x+(y<<1),now+1);
}
int main(){
	int cas; scanf("%d",&cas);
	while (cas--){
		int i;
		scanf("%lld%lld",&A,&B);
		if (A>B) swap(A,B);
		f[1]=f[2]=1;
		for (i=3; ; i++){
			f[i]=f[i-1]+f[i-2];
			if (f[i]>A || f[i]+f[i-1]>B) break;
		}
		n=i-2; ans=n==1?min(A,B):0;
		solve(1,1,0);
		swap(A,B); solve(1,1,0);
		printf("%d %lld\n",n,ans%1000000007);
	}
	return 0;
}