#include <cstdio>
#include <cctype>
#include <algorithm>
#include <cassert>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;

typedef long long LL;
const int p=1000000007;
int t,tot,ans;
LL n,m,f[100][2];

LL getLL()
{
    char ch;
    while (!isdigit(ch=getchar()));
    LL x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return x;
}

void dfs(LL x,LL y,int k)
{
    if (!k)
    {
        ans=(ans+(m-x-y)/y+1)%p;
        if (n>=x+y)
            ans=(ans+(n-x-y)/y+1)%p;
        return;
    }
    for (LL z=x+y*((!x)+1); z<=m; z+=y)
    {
        //assert(y*f[k][0]+z*f[k][1]>0 && y*f[k-1][0]+z*f[k-1][1]>0);
        if (y*f[k][0]+z*f[k][1]>m || y*f[k-1][0]+z*f[k-1][1]>n)
            break;
        if (z!=x+y*((!x)+1) && k>1)
            dfs(y*f[k-2][0]+z*f[k-2][1],y*f[k-1][0]+z*f[k-1][1],0);
        else
            dfs(y,z,k-1);
    }
}

void solve()
{
    n=getLL(),m=getLL();
    if (n>m)
        swap(n,m);
    for (tot=0; f[tot][1]<=n && f[tot+1][1]<=m; ++tot);
    if (--tot<=1)
    {
        printf("1 %d\n",n%p*m%p);
        return;
    }
    ans=0,dfs(0,1,tot-1);
    printf("%d %d\n",tot,ans);
}

int main()
{
    f[0][1]=f[1][0]=f[1][1]=1;
    repu(i,2,90)
        f[i][0]=f[i-1][0]+f[i-2][0],f[i][1]=f[i-1][1]+f[i-2][1];
    for (scanf("%d",&t); t--; solve());
    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 <cstdio>
#include <cctype>
#include <algorithm>
#include <cassert>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;

typedef long long LL;
const int p=1000000007;
int t,tot,ans;
LL n,m,f[100][2];

LL getLL()
{
    char ch;
    while (!isdigit(ch=getchar()));
    LL x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return x;
}

void dfs(LL x,LL y,int k)
{
    if (!k)
    {
        ans=(ans+(m-x-y)/y+1)%p;
        if (n>=x+y)
            ans=(ans+(n-x-y)/y+1)%p;
        return;
    }
    for (LL z=x+y*((!x)+1); z<=m; z+=y)
    {
        //assert(y*f[k][0]+z*f[k][1]>0 && y*f[k-1][0]+z*f[k-1][1]>0);
        if (y*f[k][0]+z*f[k][1]>m || y*f[k-1][0]+z*f[k-1][1]>n)
            break;
        if (z!=x+y*((!x)+1) && k>1)
            dfs(y*f[k-2][0]+z*f[k-2][1],y*f[k-1][0]+z*f[k-1][1],0);
        else
            dfs(y,z,k-1);
    }
}

void solve()
{
    n=getLL(),m=getLL();
    if (n>m)
        swap(n,m);
    for (tot=0; f[tot][1]<=n && f[tot+1][1]<=m; ++tot);
    if (--tot<=1)
    {
        printf("1 %d\n",n%p*m%p);
        return;
    }
    ans=0,dfs(0,1,tot-1);
    printf("%d %d\n",tot,ans);
}

int main()
{
    f[0][1]=f[1][0]=f[1][1]=1;
    repu(i,2,90)
        f[i][0]=f[i-1][0]+f[i-2][0],f[i][1]=f[i-1][1]+f[i-2][1];
    for (scanf("%d",&t); t--; solve());
    return 0;
}
```