#include <bits/stdc++.h>

#define x first
#define y second
#define pb push_back
#define mp make_pair
#define SZ(S) S.size()
#define vi vector<int>
#define pr pair<int,int>
#define REP(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define FOR(i,a,b) for(int i=(int)(a);i<=(int)(b);++i)
#define FORD(i,b,a) for(int i=(int)(b);i>=(int)(a);--i)
#define CLR(A,x) memset(A,x,sizeof(A))
#define ALL(S) S.begin(),S.end()

#define OUT(a,l,r) cout<<#a": ";FOR(i,l,r)cout<<a[i]<<" ";cout<<endl
#define PRT(a) cout<<"--> "#a" "<<a<<endl

using namespace std;
typedef unsigned long long ll;

const int N=90,INF=0x3f3f3f3f;
const ll P=1e9+7;

template<typename T>
inline void read(T& x,int f=1,char s=getchar()) {
	for(x=0;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1; 
	while(s<='9'&&s>='0') x=x*10+s-'0',s=getchar();x*=f;
}

ll f[N];
int w1[]={0,1,0,-1};
int w2[]={-1,0,1,0};

int main() {
	f[0]=f[1]=1;
	REP(i,2,N) f[i]=f[i-1]+f[i-2];
	
	ll T,x,y;
	read(T);
	while(T--) {
		read(x),read(y);
		if(y>x) swap(x,y);
		int now=1;
		while(f[now+2]<=x&&f[now+1]<=y) now++;
		
		if(now==1) {
			printf("%d %lld\n",now,(x*y)%P);
			continue;
		}
		
		ll a=f[now+1],b=f[now];
		ll ans=0;
		if(x>=a&&y>=b) ans=(ans+1+(x-a)/b)%P;
		if(y>=a&&x>=b) ans=(ans+1+(y-a)/b)%P;
		
		int t=0;
		FORD(i,now,0) {
			t=(t+1)%4;
			a+=f[i]*w1[t],b+=f[i]*w2[t];
			if(i!=now) {
				if(x>=a&&y>=b) ans=(ans+1+(x-a)/b)%P;
				if(y>=a&&x>=b) ans=(ans+1+(y-a)/b)%P;
			}
		}
		printf("%d %lld\n",now,ans);
	}
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 x first
#define y second
#define pb push_back
#define mp make_pair
#define SZ(S) S.size()
#define vi vector<int>
#define pr pair<int,int>
#define REP(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define FOR(i,a,b) for(int i=(int)(a);i<=(int)(b);++i)
#define FORD(i,b,a) for(int i=(int)(b);i>=(int)(a);--i)
#define CLR(A,x) memset(A,x,sizeof(A))
#define ALL(S) S.begin(),S.end()

#define OUT(a,l,r) cout<<#a": ";FOR(i,l,r)cout<<a[i]<<" ";cout<<endl
#define PRT(a) cout<<"--> "#a" "<<a<<endl

using namespace std;
typedef unsigned long long ll;

const int N=90,INF=0x3f3f3f3f;
const ll P=1e9+7;

template<typename T>
inline void read(T& x,int f=1,char s=getchar()) {
	for(x=0;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1; 
	while(s<='9'&&s>='0') x=x*10+s-'0',s=getchar();x*=f;
}

ll f[N];
int w1[]={0,1,0,-1};
int w2[]={-1,0,1,0};

int main() {
	f[0]=f[1]=1;
	REP(i,2,N) f[i]=f[i-1]+f[i-2];
	
	ll T,x,y;
	read(T);
	while(T--) {
		read(x),read(y);
		if(y>x) swap(x,y);
		int now=1;
		while(f[now+2]<=x&&f[now+1]<=y) now++;
		
		if(now==1) {
			printf("%d %lld\n",now,(x*y)%P);
			continue;
		}
		
		ll a=f[now+1],b=f[now];
		ll ans=0;
		if(x>=a&&y>=b) ans=(ans+1+(x-a)/b)%P;
		if(y>=a&&x>=b) ans=(ans+1+(y-a)/b)%P;
		
		int t=0;
		FORD(i,now,0) {
			t=(t+1)%4;
			a+=f[i]*w1[t],b+=f[i]*w2[t];
			if(i!=now) {
				if(x>=a&&y>=b) ans=(ans+1+(x-a)/b)%P;
				if(y>=a&&x>=b) ans=(ans+1+(y-a)/b)%P;
			}
		}
		printf("%d %lld\n",now,ans);
	}
	return 0;
}
```