#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>

using namespace std;
typedef long long LL;
typedef pair<LL,LL> mp;
const int maxn = 105;
const int N = 90;
const int mod = 1e9+7;
vector<mp> vec[maxn];

#define pb push_back
#define fi first
#define se second

LL fib[maxn];int Q;
void dfs(int stp,LL a,LL b)
{
	vec[stp].pb(mp(a,b));
	if (stp>=86) return ;
	LL num=fib[stp+3]+fib[stp];
	if (a+b<=num) dfs(stp+1,b,a+b);
	if (a+2*b<=num) dfs(stp+1,b,a+2*b);
}
int main()
{
	#ifdef Amberframe
		freopen("agc015f.in","r",stdin);
		freopen("agc015f.out","w",stdout);
	#endif
	fib[0]=1;fib[1]=1;
	for (int i=2;i<=N;i++)
		fib[i]=fib[i-1]+fib[i-2];
	dfs(1,1,2);
	dfs(1,1,3);
	
	scanf("%d",&Q);
	for (int i=1;i<=Q;i++)
	{
		LL x,y;scanf("%lld %lld",&x,&y);LL ans=0;
		if (x>y) swap(x,y);
		int k=1;while (x>=fib[k+1]&&y>=fib[k+2]) ++k;
		for (unsigned j=0;j<vec[k-1].size();j++)
		{
			if (vec[k-1][j].se<=x)
				ans+=max(0ll,(y-vec[k-1][j].fi)/vec[k-1][j].se);
			if (vec[k-1][j].se<=y)
				ans+=max(0ll,(x-vec[k-1][j].fi)/vec[k-1][j].se);
		}
		printf("%d %d\n",k,k==1?(LL)x*y%mod:ans%mod);
	}
	return 0;
}
### Prompt

Your challenge is to write 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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>

using namespace std;
typedef long long LL;
typedef pair<LL,LL> mp;
const int maxn = 105;
const int N = 90;
const int mod = 1e9+7;
vector<mp> vec[maxn];

#define pb push_back
#define fi first
#define se second

LL fib[maxn];int Q;
void dfs(int stp,LL a,LL b)
{
	vec[stp].pb(mp(a,b));
	if (stp>=86) return ;
	LL num=fib[stp+3]+fib[stp];
	if (a+b<=num) dfs(stp+1,b,a+b);
	if (a+2*b<=num) dfs(stp+1,b,a+2*b);
}
int main()
{
	#ifdef Amberframe
		freopen("agc015f.in","r",stdin);
		freopen("agc015f.out","w",stdout);
	#endif
	fib[0]=1;fib[1]=1;
	for (int i=2;i<=N;i++)
		fib[i]=fib[i-1]+fib[i-2];
	dfs(1,1,2);
	dfs(1,1,3);
	
	scanf("%d",&Q);
	for (int i=1;i<=Q;i++)
	{
		LL x,y;scanf("%lld %lld",&x,&y);LL ans=0;
		if (x>y) swap(x,y);
		int k=1;while (x>=fib[k+1]&&y>=fib[k+2]) ++k;
		for (unsigned j=0;j<vec[k-1].size();j++)
		{
			if (vec[k-1][j].se<=x)
				ans+=max(0ll,(y-vec[k-1][j].fi)/vec[k-1][j].se);
			if (vec[k-1][j].se<=y)
				ans+=max(0ll,(x-vec[k-1][j].fi)/vec[k-1][j].se);
		}
		printf("%d %d\n",k,k==1?(LL)x*y%mod:ans%mod);
	}
	return 0;
}
```