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>
using namespace std;
#define N 90
#define mod 1000000007

int n;
long long f[N];
vector<pair<long long,long long> > q[N];

long long read(){
	long long x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}
 
int main(){
	f[0]=f[1]=1;
	for (int i=2;i<N;i++) f[i]=f[i-2]+f[i-1];
	q[1].push_back(make_pair(1,2));
	q[1].push_back(make_pair(1,3));
	for (int i=2;i<N-2;i++)
		for (int j=0;j<(int)q[i-1].size();j++){
			long long x=q[i-1][j].second,y=q[i-1][j].first;
			for (y+=x;y<=f[i+2];y+=x) q[i].push_back(make_pair(x,y));
		}
	for (int t=read();t;t--){
		long long x=read(),y=read(),ans=1,cnt=0;
		if (x>y) swap(x,y);
		while (f[ans+1]<=x&&f[ans+2]<=y) ans++;
		if (ans==1){
			printf ("1 %lld\n",1ll*x*y%mod);
			continue;
		}
		for (int i=0;i<(int)q[ans-1].size();i++){
			long long tx=q[ans-1][i].first,ty=q[ans-1][i].second;
			if (x>=ty&&y>=tx+ty) cnt=(cnt+(y-tx)/ty)%mod;
			if (y>=ty&&x>=tx+ty) cnt=(cnt+(x-tx)/ty)%mod;
		}
		printf("%lld %lld\n",ans,cnt);
	}
	return 0;
}