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 <cstdio>
#include <cstring>
#include <vector>
#define mp make_pair
using namespace std;
typedef long long ll;
typedef pair<ll,ll> data;
const int mod=1e9+7;
vector<data> a[205];
ll F[205];
const ll inf=1e18+10;
int Q;
int main(){
	F[0]=F[1]=1;
	for(int i=2;i<=200;i++) F[i]=F[i-1]+F[i-2];
	a[1].push_back(mp(1,2));
	a[1].push_back(mp(1,3));
	a[1].push_back(mp(1,4));
	for(int i=1;i<=200;i++){
		for(int p=0,S=a[i].size();p<S;p++){
			ll x=a[i][p].first,y=a[i][p].second;
			x+=y;
			while(x<=F[i+3]+F[i]){
				a[i+1].push_back(mp(y,x));
				x+=y;
			}
		}
	}
	scanf("%d",&Q);
	ll n,m;
	while(Q--){
		scanf("%lld%lld",&n,&m);
		if(n>m) swap(n,m);
		int ans=1;
		while(F[ans+1]<=n&&F[ans+2]<=m) ans++;
		if(ans<=1){
			printf("%d %lld\n",ans,n*m%mod);
			continue;
		}
		ll num=0;
		for(int i=0,S=a[ans-1].size();i<S;i++){
			ll x=a[ans-1][i].first,y=a[ans-1][i].second;
			if(y<=n) num=(num+(m-x)/y)%mod;
			if(y<=m) num=(num+(n-x)/y)%mod;
		}
		printf("%d %lld\n",ans,num);
	}
	return 0;
}