#include<cstdio>
#include<algorithm>
#include<vector>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pi;
const int mx=90+10,mo=1000000007;
const ll inf=1000000000000000000;
ll fib[mx];
vector<pi> a[mx];
vector<pi>::iterator it;
ll x,y,n,m,num;
int i,j,k,l,t,tot,ca,ans;
int main(){
    fib[0]=fib[1]=1;
    tot=1;
    while (1){
        fib[tot+1]=fib[tot]+fib[tot-1];
        if (fib[tot+1]>inf) break;
        tot++;
    }
    a[1].pb(mp(1,2));a[1].pb(mp(1,3));a[1].pb(mp(1,4));
    fo(i,1,tot-3){
        it=a[i].begin();
        while (it!=a[i].end()){
            x=(*it).xx;y=(*it).yy;
            x+=y;
            while (x<=fib[i+3]){
                a[i+1].pb(mp(y,x));
                x+=y;
            }
            it++;
        }
    }
    scanf("%d",&ca);
    while (ca--){
        scanf("%lld%lld",&n,&m);
        if (n>m) swap(n,m);
        ans=1;
        while (fib[ans+1]<=n&&fib[ans+2]<=m) ans++;
        printf("%d ",ans);
        if (ans==1){
            printf("%lld\n",n*m%mo);
            continue;
        }
        else{
            num=0;
            it=a[ans-1].begin();
            while (it!=a[ans-1].end()){
                x=(*it).xx;y=(*it).yy;
                if (y<=n) (num+=(m-x)/y)%=mo;
                if (y<=m) (num+=(n-x)/y)%=mo;
                it++;
            }
            printf("%d\n",num);
        }
    }
}
### 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<cstdio>
#include<algorithm>
#include<vector>
#define xx first
#define yy second
#define mp make_pair
#define pb push_back
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pi;
const int mx=90+10,mo=1000000007;
const ll inf=1000000000000000000;
ll fib[mx];
vector<pi> a[mx];
vector<pi>::iterator it;
ll x,y,n,m,num;
int i,j,k,l,t,tot,ca,ans;
int main(){
    fib[0]=fib[1]=1;
    tot=1;
    while (1){
        fib[tot+1]=fib[tot]+fib[tot-1];
        if (fib[tot+1]>inf) break;
        tot++;
    }
    a[1].pb(mp(1,2));a[1].pb(mp(1,3));a[1].pb(mp(1,4));
    fo(i,1,tot-3){
        it=a[i].begin();
        while (it!=a[i].end()){
            x=(*it).xx;y=(*it).yy;
            x+=y;
            while (x<=fib[i+3]){
                a[i+1].pb(mp(y,x));
                x+=y;
            }
            it++;
        }
    }
    scanf("%d",&ca);
    while (ca--){
        scanf("%lld%lld",&n,&m);
        if (n>m) swap(n,m);
        ans=1;
        while (fib[ans+1]<=n&&fib[ans+2]<=m) ans++;
        printf("%d ",ans);
        if (ans==1){
            printf("%lld\n",n*m%mo);
            continue;
        }
        else{
            num=0;
            it=a[ans-1].begin();
            while (it!=a[ans-1].end()){
                x=(*it).xx;y=(*it).yy;
                if (y<=n) (num+=(m-x)/y)%=mo;
                if (y<=m) (num+=(n-x)/y)%=mo;
                it++;
            }
            printf("%d\n",num);
        }
    }
}
```