//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ld long double
#define al(a) (a).begin(),(a).end()
#define mk make_pair
#define check cout<<"?"<<endl;

ll MOD=1000000007;
ll mod=998244353;
int inf=1000001000;
ll INF=1e18+5;

int main(){
    vector<ll> a(88,0);
    a[0]=a[1]=1;
    for(int i=2;i<88;i++){
        a[i]=a[i-1]+a[i-2];
    }
    ll q; cin>>q;
    rep(_,q){
        ll x,y,ans,cnt=0; cin>>x>>y;
        if(x<y) swap(x,y);
        for(int i=85;i>=1;i--){
            if(x==1 && y==1){
                ans=1; break;
            }
            if(a[i+1]<=x && a[i]<=y){
                ans=i; break;
            }
        }
        if(ans==1){
            for(int i=1;i<=y;i++) cnt=(cnt+x/i)%MOD;
            if(x==2 && y==2) cnt=(cnt+1)%MOD;
        }
        else{
            ll X,Y;
            for(int i=0;i<ans;i++){
                if(i==0){
                    X=a[ans]; Y=a[ans+1];
                }
                else{
                    X=a[ans]+a[i]*a[ans-i-1];
                    Y=a[ans+1]+a[i]*a[ans-i];
                }
                if(X<=y) cnt=(cnt+(x-Y+X)/X)%MOD;
                if(X<=x) cnt=(cnt+(y-Y+X)/X)%MOD;
            }
        }
        printf("%lld %lld\n",ans,cnt);
    }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ld long double
#define al(a) (a).begin(),(a).end()
#define mk make_pair
#define check cout<<"?"<<endl;

ll MOD=1000000007;
ll mod=998244353;
int inf=1000001000;
ll INF=1e18+5;

int main(){
    vector<ll> a(88,0);
    a[0]=a[1]=1;
    for(int i=2;i<88;i++){
        a[i]=a[i-1]+a[i-2];
    }
    ll q; cin>>q;
    rep(_,q){
        ll x,y,ans,cnt=0; cin>>x>>y;
        if(x<y) swap(x,y);
        for(int i=85;i>=1;i--){
            if(x==1 && y==1){
                ans=1; break;
            }
            if(a[i+1]<=x && a[i]<=y){
                ans=i; break;
            }
        }
        if(ans==1){
            for(int i=1;i<=y;i++) cnt=(cnt+x/i)%MOD;
            if(x==2 && y==2) cnt=(cnt+1)%MOD;
        }
        else{
            ll X,Y;
            for(int i=0;i<ans;i++){
                if(i==0){
                    X=a[ans]; Y=a[ans+1];
                }
                else{
                    X=a[ans]+a[i]*a[ans-i-1];
                    Y=a[ans+1]+a[i]*a[ans-i];
                }
                if(X<=y) cnt=(cnt+(x-Y+X)/X)%MOD;
                if(X<=x) cnt=(cnt+(y-Y+X)/X)%MOD;
            }
        }
        printf("%lld %lld\n",ans,cnt);
    }
}

```