#include<bits/stdc++.h>
#define maxn 110
#define all 100
#define p 1000000007
#define mk make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
template<typename T> inline void read(T &x)
{
    x=0;char c=getchar();bool flag=false;
    while(!isdigit(c)){if(c=='-')flag=true;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    if(flag)x=-x;
}
int q;
ll f[maxn];
vector<pair<ll,ll> > ve[maxn];
int main()
{
    f[0]=f[1]=1;
    for(int i=2;i<=all;++i) f[i]=f[i-1]+f[i-2];
    ve[1].push_back(mk(1,2)),ve[1].push_back(mk(1,3)),ve[1].push_back(mk(1,4));
    for(int i=2;i<=all;++i)
    {
        for(int j=0;j<ve[i-1].size();++j)
        {
            ll x=ve[i-1][j].se,y=ve[i-1][j].fi+x;
            while(y<=f[i+2]+f[i-1]) ve[i].push_back(mk(x,y)),y+=x;
        }
    }
    read(q);
    for(int i=1;i<=q;++i)
    {
        ll n,m,mx=1,v=0;
        read(n),read(m);
        if(n>m) swap(n,m);
        while(f[mx+1]<=n&&f[mx+2]<=m) mx++;
        printf("%lld ",mx);
        if(mx==1)
        {
            printf("%lld\n",n%p*m%p);
            continue;
        }
        for(int j=0;j<ve[mx-1].size();++j)
        {
            ll x=ve[mx-1][j].fi,y=ve[mx-1][j].se;
            if(y<=n) v=(v+(m-x)/y)%p;
            if(y<=m) v=(v+(n-x)/y)%p;
        }
        printf("%lld\n",v);
    }
    return 0;
}
### Prompt

Generate a cpp solution to 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<bits/stdc++.h>
#define maxn 110
#define all 100
#define p 1000000007
#define mk make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
template<typename T> inline void read(T &x)
{
    x=0;char c=getchar();bool flag=false;
    while(!isdigit(c)){if(c=='-')flag=true;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    if(flag)x=-x;
}
int q;
ll f[maxn];
vector<pair<ll,ll> > ve[maxn];
int main()
{
    f[0]=f[1]=1;
    for(int i=2;i<=all;++i) f[i]=f[i-1]+f[i-2];
    ve[1].push_back(mk(1,2)),ve[1].push_back(mk(1,3)),ve[1].push_back(mk(1,4));
    for(int i=2;i<=all;++i)
    {
        for(int j=0;j<ve[i-1].size();++j)
        {
            ll x=ve[i-1][j].se,y=ve[i-1][j].fi+x;
            while(y<=f[i+2]+f[i-1]) ve[i].push_back(mk(x,y)),y+=x;
        }
    }
    read(q);
    for(int i=1;i<=q;++i)
    {
        ll n,m,mx=1,v=0;
        read(n),read(m);
        if(n>m) swap(n,m);
        while(f[mx+1]<=n&&f[mx+2]<=m) mx++;
        printf("%lld ",mx);
        if(mx==1)
        {
            printf("%lld\n",n%p*m%p);
            continue;
        }
        for(int j=0;j<ve[mx-1].size();++j)
        {
            ll x=ve[mx-1][j].fi,y=ve[mx-1][j].se;
            if(y<=n) v=(v+(m-x)/y)%p;
            if(y<=m) v=(v+(n-x)/y)%p;
        }
        printf("%lld\n",v);
    }
    return 0;
}
```