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 fir first
#define sec second
#define db double
#define ll long long
#define ev eg[i].v
#define pb push_back
#define mod 1000000007
#define pir pair<ll,ll>
#define Rep(i,l,r) for(int i=(l);i<=(r);++i)
#define RepD(i,r,l) for(int i=(r);i>=(l);--i)
#define RepG(i,x) for(int i=hd[x];i;i=eg[i].nxt)
using namespace std;

const int N=95;

ll f[N];
vector<pir >a[N];

int main(){
	ll q,n,m;
	f[0]=f[1]=1;
	Rep(i,2,N-5) f[i]=f[i-1]+f[i-2];
	a[1].pb(pir(1,2)),a[1].pb(pir(1,3)),a[1].pb(pir(1,4));
	Rep(i,1,N-5)
		for(int j=0;j<a[i].size();++j){
			ll x=a[i][j].fir,y=a[i][j].sec;x+=y;
			while(x<=f[i+3]+f[i]) a[i+1].pb(pir(y,x)),x+=y;
		}
	scanf("%lld",&q);
	while(q--){
		ll ans=1,res=0;
		scanf("%lld%lld",&n,&m);
		if(n>m) swap(n,m);
		while(f[ans+1]<=n&&f[ans+2]<=m) ans++;
		if(ans==1) res=n*m%mod;
		else for(int i=0;i<a[ans-1].size();++i){
			ll x=a[ans-1][i].fir,y=a[ans-1][i].sec;
			if(y<=n) res=(res+(m-x)/y)%mod;
			if(y<=m) res=(res+(n-x)/y)%mod;
		}
		printf("%lld %lld\n",ans,res);
	}
	return 0;
}