#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
  ll mod=1e9+7;
  ll vec[105];
  vec[0]=1,vec[1]=1;
  for(int i=2;i<105;i++){
    vec[i]=vec[i-1]+vec[i-2];
  }
  ll pair[101][101][2];
  for(int i=2;i<101;i++){
    for(int j=0;j<i;j++){
      if(j==0){
        pair[i][j][0]=vec[i];
        pair[i][j][1]=vec[i+1];
        continue;
      }
      if(j==1){
        pair[i][j][0]=vec[i+1];
        pair[i][j][1]=2*vec[i+1]-vec[i];
        continue;
      }
      pair[i][j][0]=pair[i-1][j-1][1];
      pair[i][j][1]=pair[i-1][j-1][1]+pair[i-1][j-1][0];
    }
  }
  int q;
  cin>>q;
  ll x,y,z=1;
  ll w=0;
  ll o,p,r;
  for(int i=0;i<q;i++){
    cin>>x>>y;
    if(x>y){
      swap(x,y);
    }
    if(y==1){
      cout<<"1 1"<<endl;
      continue;
    }
    for(int j=0;j<98;j++){
      if(vec[j]>x||vec[j+1]>y){
        z=j-1;
        break;
      }
    }
    cout<<z<<" ";
    if(z==1){
      cout<<(x%mod*y%mod)%mod<<endl;
      continue;
    }
    w=0;
    for(int j=0;j<z;j++){
      o=pair[z][j][0];
      p=pair[z][j][1];
      if(o>x||p>y){
        continue;
      }
      r=y-p;
      r/=o;
      w+=r+1;
      w%=mod;
      if(x>=p){
        r=x-p;
        r/=o;
        w+=r+1;
        w%=mod;
      }
    }
    cout<<w<<endl;
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
  ll mod=1e9+7;
  ll vec[105];
  vec[0]=1,vec[1]=1;
  for(int i=2;i<105;i++){
    vec[i]=vec[i-1]+vec[i-2];
  }
  ll pair[101][101][2];
  for(int i=2;i<101;i++){
    for(int j=0;j<i;j++){
      if(j==0){
        pair[i][j][0]=vec[i];
        pair[i][j][1]=vec[i+1];
        continue;
      }
      if(j==1){
        pair[i][j][0]=vec[i+1];
        pair[i][j][1]=2*vec[i+1]-vec[i];
        continue;
      }
      pair[i][j][0]=pair[i-1][j-1][1];
      pair[i][j][1]=pair[i-1][j-1][1]+pair[i-1][j-1][0];
    }
  }
  int q;
  cin>>q;
  ll x,y,z=1;
  ll w=0;
  ll o,p,r;
  for(int i=0;i<q;i++){
    cin>>x>>y;
    if(x>y){
      swap(x,y);
    }
    if(y==1){
      cout<<"1 1"<<endl;
      continue;
    }
    for(int j=0;j<98;j++){
      if(vec[j]>x||vec[j+1]>y){
        z=j-1;
        break;
      }
    }
    cout<<z<<" ";
    if(z==1){
      cout<<(x%mod*y%mod)%mod<<endl;
      continue;
    }
    w=0;
    for(int j=0;j<z;j++){
      o=pair[z][j][0];
      p=pair[z][j][1];
      if(o>x||p>y){
        continue;
      }
      r=y-p;
      r/=o;
      w+=r+1;
      w%=mod;
      if(x>=p){
        r=x-p;
        r/=o;
        w+=r+1;
        w%=mod;
      }
    }
    cout<<w<<endl;
  }
}

```