#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long LL;
const int mo=1e9+7;
const int K=87;
int q;
LL fib[K+10],cfib[K+10][K+10];
int Sum(int x,int y) {
    x+=y;
    return (x>=mo)?x-mo:x;
}
int Sub(int x,int y) {
    x-=y;
    return (x<0)?x+mo:x;
}
int Mul(int x,int y) {
    return (LL)x*y%mo;
}
void Prework() {
    fib[0]=0;
    fib[1]=1;
    fib[2]=2;
    for (int i=3;i<=K;i++)
        fib[i]=fib[i-1]+fib[i-2];
    for (int i=1;i<=K;i++) cfib[1][i]=fib[i];
    for (int i=2;i<=K;i++) {
        for (int j=1;j<i;j++) cfib[i][j]=fib[j];
        cfib[i][i]=cfib[i][i-1]*2+cfib[i][i-2];
        if (i==2) cfib[i][i]=cfib[i][i-1]*3;
        for (int j=i+1;j<=K;j++)
            cfib[i][j]=cfib[i][j-1]+cfib[i][j-2];
    }
}
void Calc(LL l1,LL l2) {
    if (l1>l2) swap(l1,l2);
    if (l1==1&&l2==1) {
        printf("1 1\n");
        return;
    }
    int cnt=0;
    for (int i=2;i<=K;i++)
        if (fib[i-1]<=l1&&fib[i]<=l2) cnt++;
    if (cnt==1) {
        printf("1 %d\n",Mul(l1%mo,l2%mo));
        return;
    }
    int sum=0;
    for (int i=1;i<=cnt;i++) {
        if (cfib[i][cnt]>l1||cfib[i][cnt+1]>l2) continue;
        sum=Sum(sum,((l2-cfib[i][cnt+1])/cfib[i][cnt]+1)%mo);
        if (cfib[i][cnt+1]>l1||cfib[i][cnt]>l2) continue;
        sum=Sum(sum,((l1-cfib[i][cnt+1])/cfib[i][cnt]+1)%mo);
    }
    printf("%d %d\n",cnt,sum);
}
int main() {
    Prework();
    scanf("%d",&q);
    while (q--) {
        LL l1,l2;
        scanf("%lld%lld",&l1,&l2);
        Calc(l1,l2);
    }
    return 0;
}
### 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 <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long LL;
const int mo=1e9+7;
const int K=87;
int q;
LL fib[K+10],cfib[K+10][K+10];
int Sum(int x,int y) {
    x+=y;
    return (x>=mo)?x-mo:x;
}
int Sub(int x,int y) {
    x-=y;
    return (x<0)?x+mo:x;
}
int Mul(int x,int y) {
    return (LL)x*y%mo;
}
void Prework() {
    fib[0]=0;
    fib[1]=1;
    fib[2]=2;
    for (int i=3;i<=K;i++)
        fib[i]=fib[i-1]+fib[i-2];
    for (int i=1;i<=K;i++) cfib[1][i]=fib[i];
    for (int i=2;i<=K;i++) {
        for (int j=1;j<i;j++) cfib[i][j]=fib[j];
        cfib[i][i]=cfib[i][i-1]*2+cfib[i][i-2];
        if (i==2) cfib[i][i]=cfib[i][i-1]*3;
        for (int j=i+1;j<=K;j++)
            cfib[i][j]=cfib[i][j-1]+cfib[i][j-2];
    }
}
void Calc(LL l1,LL l2) {
    if (l1>l2) swap(l1,l2);
    if (l1==1&&l2==1) {
        printf("1 1\n");
        return;
    }
    int cnt=0;
    for (int i=2;i<=K;i++)
        if (fib[i-1]<=l1&&fib[i]<=l2) cnt++;
    if (cnt==1) {
        printf("1 %d\n",Mul(l1%mo,l2%mo));
        return;
    }
    int sum=0;
    for (int i=1;i<=cnt;i++) {
        if (cfib[i][cnt]>l1||cfib[i][cnt+1]>l2) continue;
        sum=Sum(sum,((l2-cfib[i][cnt+1])/cfib[i][cnt]+1)%mo);
        if (cfib[i][cnt+1]>l1||cfib[i][cnt]>l2) continue;
        sum=Sum(sum,((l1-cfib[i][cnt+1])/cfib[i][cnt]+1)%mo);
    }
    printf("%d %d\n",cnt,sum);
}
int main() {
    Prework();
    scanf("%d",&q);
    while (q--) {
        LL l1,l2;
        scanf("%lld%lld",&l1,&l2);
        Calc(l1,l2);
    }
    return 0;
}
```