#include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<cassert>
#define PB push_back
#define MP make_pair
#define sz(v) (in((v).size()))
#define forn(i,n) for(in i=0;i<(n);++i)
#define forv(i,v) forn(i,sz(v))
#define fors(i,s) for(auto i=(s).begin();i!=(s).end();++i)
#define all(v) (v).begin(),(v).end()
using namespace std;
typedef long long in;
typedef vector<in> VI;
typedef vector<VI> VVI;
in ma,mb;
in sc;
in sm;
in ts,tx,ty;
const in mdl=1000000007;
VI amax,amay,ambx,amby;
void sup(in x, in y, in s){
  if(amax[s]*x+amay[s]*y>ma)
    return;
  if(ambx[s]*x+amby[s]*y>mb)
    return;
  if(s==sc){
    sm+=(ma-x)/y+1;
    if(x<=mb){
      ++sm;
    }
    sm%=mdl;
    return;
  }
  sup(x+y,y,s);
  sup(x+y,x,s+1);
}
void slv(){
  if(mb>ma)
    swap(mb,ma);
  if(mb==1){
    cout<<1<<" "<<ma%mdl<<"\n";
    return;
  }
  if(mb==2 && ma==2){
    cout<<1<<" "<<4<<"\n";
    return;
  }
  in x=2;
  in y=1;
  sc=2;
  while(x+y<=ma && x<=mb){
    y+=x;
    swap(x,y);
    ++sc;
  }
  amax[sc]=1;
  amay[sc]=0;
  ambx[sc]=0;
  amby[sc]=1;
  for(in i=sc-1;i>=2;--i){
    amax[i]=amax[i+1]+amay[i+1];
    amay[i]=amax[i+1];
    ambx[i]=ambx[i+1]+amby[i+1];
    amby[i]=ambx[i+1];
  }
  cout<<sc-1<<" ";
  sm=0;
  sup(2,1,2);
  cout<<sm<<"\n";
}
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  amax=amay=ambx=amby=VI(200);
  in q;
  cin>>q;
  forn(z,q){
    cin>>ma>>mb;
    slv();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve the following 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

### Response

```cpp
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<cassert>
#define PB push_back
#define MP make_pair
#define sz(v) (in((v).size()))
#define forn(i,n) for(in i=0;i<(n);++i)
#define forv(i,v) forn(i,sz(v))
#define fors(i,s) for(auto i=(s).begin();i!=(s).end();++i)
#define all(v) (v).begin(),(v).end()
using namespace std;
typedef long long in;
typedef vector<in> VI;
typedef vector<VI> VVI;
in ma,mb;
in sc;
in sm;
in ts,tx,ty;
const in mdl=1000000007;
VI amax,amay,ambx,amby;
void sup(in x, in y, in s){
  if(amax[s]*x+amay[s]*y>ma)
    return;
  if(ambx[s]*x+amby[s]*y>mb)
    return;
  if(s==sc){
    sm+=(ma-x)/y+1;
    if(x<=mb){
      ++sm;
    }
    sm%=mdl;
    return;
  }
  sup(x+y,y,s);
  sup(x+y,x,s+1);
}
void slv(){
  if(mb>ma)
    swap(mb,ma);
  if(mb==1){
    cout<<1<<" "<<ma%mdl<<"\n";
    return;
  }
  if(mb==2 && ma==2){
    cout<<1<<" "<<4<<"\n";
    return;
  }
  in x=2;
  in y=1;
  sc=2;
  while(x+y<=ma && x<=mb){
    y+=x;
    swap(x,y);
    ++sc;
  }
  amax[sc]=1;
  amay[sc]=0;
  ambx[sc]=0;
  amby[sc]=1;
  for(in i=sc-1;i>=2;--i){
    amax[i]=amax[i+1]+amay[i+1];
    amay[i]=amax[i+1];
    ambx[i]=ambx[i+1]+amby[i+1];
    amby[i]=ambx[i+1];
  }
  cout<<sc-1<<" ";
  sm=0;
  sup(2,1,2);
  cout<<sm<<"\n";
}
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  amax=amay=ambx=amby=VI(200);
  in q;
  cin>>q;
  forn(z,q){
    cin>>ma>>mb;
    slv();
  }
  return 0;
}

```