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;

inline long long read()
{
	long long x=0,f=1,c=getchar();
	while(c<48) c=='-'&&(f=-1),c=getchar();
	while(c>47) x=x*10+c-'0',c=getchar();
	return x*f;
}

typedef long long ll;
const int mod = 1e9+7;
const int MAXN = 10005;
vector<pair<ll,ll> > v[MAXN];
ll f[MAXN];

inline void init()
{
	for(int i=(f[0]=f[1]=1,2); i<=90; i++)
		f[i]=f[i-1]+f[i-2];
	v[1].push_back(make_pair(1,2));
	v[1].push_back(make_pair(1,3));
	v[1].push_back(make_pair(1,4));
	for(int i=1; i<=90-3; i++)
		for(auto p : v[i])
		{
			ll x=p.first,y=p.second,tmp=x; x=y; y=tmp+y;
			while(y<=f[i+3]+f[i-1])
				v[i+1].push_back(make_pair(x,y)),y+=x;
		}
}

int main(int argc, char const *argv[])
{
	init();
	for(int T=read(); T; --T)
	{
		ll x=read(),y=read(),k=1,ans=0;
		if(x>y) swap(x,y);
		while(f[k+1]<=x&&f[k+2]<=y) ++k;
		printf("%lld ", k);
		if(k==1) {printf("%lld\n", x*y%mod); continue;}
		for(auto p : v[k-1])
		{
			ll a=p.first,b=p.second;
			if(b<=x) ans+=(y-a)/b%mod;
			if(b<=y) ans+=(x-a)/b%mod; ans%=mod;
		}
		printf("%lld\n", ans%mod);
	}
	return 0;
}