#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#define INF 1000000000000000000LL
#define MOD 1000000007LL

using namespace std;
typedef long long int ll;

int main()
{
	int q;
	scanf("%d", &q);
	vector<ll> fib;
	fib.push_back(0);
	fib.push_back(1);
	while(fib.back()<=INF){
		int i=fib.size();
		fib.push_back(fib[i-1]+fib[i-2]);
	}
	int max=fib.size(); // max=89
	ll prod[100][100];
	for(int k=2; k<max; k++){
		for(int i=1; i<k; i++){
			prod[k][i]=fib[i]*fib[k-i];
		}
	}

	for(int i=0; i<q; i++){
		ll x, y;
		scanf("%lld %lld", &x, &y);
		if(x>y) swap(x, y);
		if(x==1){
			printf("%d %lld\n", 1, y%MOD);
			continue;
		}else if(x==2){
			if(y==2){
				printf("%d %d\n", 1, 4);
			}else{
				printf("%d %lld\n", 2, (y-1)/2%MOD);
			}
			continue;
		}
		int n1=upper_bound(fib.begin(), fib.end(), x)-fib.begin()-1;
		int n2=upper_bound(fib.begin(), fib.end(), y)-fib.begin()-1;
		int n=min(n1, n2-1);
		ll ans=0;
		ans+=((y-fib[n+1])/fib[n]+1);
		ans%=MOD;
		for(int i=1; i<n; i++){
			ll x1=fib[n]+prod[n][i], y1=fib[n+1]+prod[n+1][i+1];
			if(x1>x || y1>y) continue;
			ans+=((y-y1)/x1+1);
			ans%=MOD;
		}
		if(x<fib[n+1] || y<fib[n]){
			printf("%d %lld\n", n-1, ans);
			continue;
		}
		ans+=((x-fib[n+1])/fib[n]+1);
		ans%=MOD;
		for(int i=1; i<n; i++){
			ll x1=fib[n]+prod[n][i], y1=fib[n+1]+prod[n+1][i+1];
			if(x1>y || y1>x) continue;
			ans+=((x-y1)/x1+1);
			ans%=MOD;
		}
		printf("%d %lld\n", n-1, ans);
	}
	return 0;
}
### Prompt

Your task is to create 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 <cstring>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#define INF 1000000000000000000LL
#define MOD 1000000007LL

using namespace std;
typedef long long int ll;

int main()
{
	int q;
	scanf("%d", &q);
	vector<ll> fib;
	fib.push_back(0);
	fib.push_back(1);
	while(fib.back()<=INF){
		int i=fib.size();
		fib.push_back(fib[i-1]+fib[i-2]);
	}
	int max=fib.size(); // max=89
	ll prod[100][100];
	for(int k=2; k<max; k++){
		for(int i=1; i<k; i++){
			prod[k][i]=fib[i]*fib[k-i];
		}
	}

	for(int i=0; i<q; i++){
		ll x, y;
		scanf("%lld %lld", &x, &y);
		if(x>y) swap(x, y);
		if(x==1){
			printf("%d %lld\n", 1, y%MOD);
			continue;
		}else if(x==2){
			if(y==2){
				printf("%d %d\n", 1, 4);
			}else{
				printf("%d %lld\n", 2, (y-1)/2%MOD);
			}
			continue;
		}
		int n1=upper_bound(fib.begin(), fib.end(), x)-fib.begin()-1;
		int n2=upper_bound(fib.begin(), fib.end(), y)-fib.begin()-1;
		int n=min(n1, n2-1);
		ll ans=0;
		ans+=((y-fib[n+1])/fib[n]+1);
		ans%=MOD;
		for(int i=1; i<n; i++){
			ll x1=fib[n]+prod[n][i], y1=fib[n+1]+prod[n+1][i+1];
			if(x1>x || y1>y) continue;
			ans+=((y-y1)/x1+1);
			ans%=MOD;
		}
		if(x<fib[n+1] || y<fib[n]){
			printf("%d %lld\n", n-1, ans);
			continue;
		}
		ans+=((x-fib[n+1])/fib[n]+1);
		ans%=MOD;
		for(int i=1; i<n; i++){
			ll x1=fib[n]+prod[n][i], y1=fib[n+1]+prod[n+1][i+1];
			if(x1>y || y1>x) continue;
			ans+=((x-y1)/x1+1);
			ans%=MOD;
		}
		printf("%d %lld\n", n-1, ans);
	}
	return 0;
}
```