#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair <LL,LL> pii;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1 : *ib++;
}
inline LL rd() {
	char ch=nc(); LL i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}
inline void W(LL x) {
	static int buf[50];
	if(!x) {putchar('0'); return;}
	if(x<0) {putchar('-'); x=-x;}
	while(x) {buf[++buf[0]]=x%10; x/=10;}
	while(buf[0]) putchar(buf[buf[0]--]+'0');
}

const int N=233, mod=1e9+7;
LL f[N],n;
vector <pii> g[N];

inline LL calc(pii v,LL X,LL Y) {
	LL ret=0;
	if(X>=v.second && Y>=v.first+v.second) ret=(ret+(Y-v.first)/v.second)%mod;
	if(Y>=v.second && X>=v.first+v.second) ret=(ret+(X-v.first)/v.second)%mod;
	return ret;
}
int main() {
	f[0]=f[n=1]=1;
	while(f[n]<=4e18) ++n, f[n]=f[n-1]+f[n-2];
	for(int i=1;i<=f[2];i++) g[0].push_back(pii(0,i));
	for(int i=0;i<n-2;i++) 
		for(auto j:g[i]) {
			LL x=j.first, y=j.second;
			x+=y;
			while(x<=f[i+3]+f[i]) {
				if(x>y) g[i+1].push_back(pii(y,x));
				x+=y;
			}
		}
	int Q=rd();
	while(Q--) {
		LL X=rd(), Y=rd();
		if(X<Y) swap(X,Y);
		LL ans=1, sum=0;
		for(;f[ans+2]<=X && f[ans+1]<=Y;++ans);	
		if(ans==1) {sum=X*Y%mod;}
		else for(auto v:g[ans-1]) (sum+=calc(v,X,Y))%=mod;
		W(ans), putchar(' '), W(sum), putchar('\n');
	}
}

### Prompt

In CPP, your task is to solve 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 <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair <LL,LL> pii;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1 : *ib++;
}
inline LL rd() {
	char ch=nc(); LL i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}
inline void W(LL x) {
	static int buf[50];
	if(!x) {putchar('0'); return;}
	if(x<0) {putchar('-'); x=-x;}
	while(x) {buf[++buf[0]]=x%10; x/=10;}
	while(buf[0]) putchar(buf[buf[0]--]+'0');
}

const int N=233, mod=1e9+7;
LL f[N],n;
vector <pii> g[N];

inline LL calc(pii v,LL X,LL Y) {
	LL ret=0;
	if(X>=v.second && Y>=v.first+v.second) ret=(ret+(Y-v.first)/v.second)%mod;
	if(Y>=v.second && X>=v.first+v.second) ret=(ret+(X-v.first)/v.second)%mod;
	return ret;
}
int main() {
	f[0]=f[n=1]=1;
	while(f[n]<=4e18) ++n, f[n]=f[n-1]+f[n-2];
	for(int i=1;i<=f[2];i++) g[0].push_back(pii(0,i));
	for(int i=0;i<n-2;i++) 
		for(auto j:g[i]) {
			LL x=j.first, y=j.second;
			x+=y;
			while(x<=f[i+3]+f[i]) {
				if(x>y) g[i+1].push_back(pii(y,x));
				x+=y;
			}
		}
	int Q=rd();
	while(Q--) {
		LL X=rd(), Y=rd();
		if(X<Y) swap(X,Y);
		LL ans=1, sum=0;
		for(;f[ans+2]<=X && f[ans+1]<=Y;++ans);	
		if(ans==1) {sum=X*Y%mod;}
		else for(auto v:g[ans-1]) (sum+=calc(v,X,Y))%=mod;
		W(ans), putchar(' '), W(sum), putchar('\n');
	}
}

```