#include<bits/stdc++.h>
#define Fst first
#define Snd second
#define RG register
#define mp make_pair
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned int UI;
typedef unsigned long long ULL;
template<typename T> inline void read(T& x) {
	char c = getchar();
	bool f = false;
	for (x = 0; !isdigit(c); c = getchar()) {
		if (c == '-') {
			f = true;
		}
	}
	for (; isdigit(c); c = getchar()) {
		x = x * 10 + c - '0';
	}
	if (f) {
		x = -x;
	}
}
template<typename T, typename... U> inline void read(T& x, U& ... y) {
	read(x), read(y...);
}
const LL MAX=4e18;
const int P=1e9+7;
int n,Q;
LL F[110];
vector<pair<LL,LL> > G[110];
//#define rua
int main() {
//	ios::sync_with_stdio(false);
#ifdef rua
	freopen("GG.in","r",stdin);
#endif
  F[0]=F[n=1]=1;
  while(F[n-1]+F[n]<=MAX) ++n,F[n]=F[n-2]+F[n-1];
  for(int i=2;i<=4;++i) G[1].push_back(mp(1,i));
  for(int i=1;i<=n-3;++i) {
    for(auto v: G[i]) {
      LL x=v.Snd,y=v.Fst+x;
      while(y<=F[i+3]+F[i]) G[i+1].push_back(mp(x,y)),y+=x;
    }
  }
  read(Q);
  while(Q--) {
    LL X,Y; read(X,Y);
    if(X>Y) swap(X,Y);
    LL t=1,res=0;
    while(F[t+1]<=X&&F[t+2]<=Y) ++t;
    if(t==1) res=X%P*Y%P;
    else {
      for(auto v: G[t-1]) {
        LL x=v.Fst,y=v.Snd;
        if(y<=X) (res+=(Y-x)/y)%=P;
        if(y<=Y) (res+=(X-x)/y)%=P;
      }
    }
    printf("%lld %lld\n",t,res);
  }
	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<bits/stdc++.h>
#define Fst first
#define Snd second
#define RG register
#define mp make_pair
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned int UI;
typedef unsigned long long ULL;
template<typename T> inline void read(T& x) {
	char c = getchar();
	bool f = false;
	for (x = 0; !isdigit(c); c = getchar()) {
		if (c == '-') {
			f = true;
		}
	}
	for (; isdigit(c); c = getchar()) {
		x = x * 10 + c - '0';
	}
	if (f) {
		x = -x;
	}
}
template<typename T, typename... U> inline void read(T& x, U& ... y) {
	read(x), read(y...);
}
const LL MAX=4e18;
const int P=1e9+7;
int n,Q;
LL F[110];
vector<pair<LL,LL> > G[110];
//#define rua
int main() {
//	ios::sync_with_stdio(false);
#ifdef rua
	freopen("GG.in","r",stdin);
#endif
  F[0]=F[n=1]=1;
  while(F[n-1]+F[n]<=MAX) ++n,F[n]=F[n-2]+F[n-1];
  for(int i=2;i<=4;++i) G[1].push_back(mp(1,i));
  for(int i=1;i<=n-3;++i) {
    for(auto v: G[i]) {
      LL x=v.Snd,y=v.Fst+x;
      while(y<=F[i+3]+F[i]) G[i+1].push_back(mp(x,y)),y+=x;
    }
  }
  read(Q);
  while(Q--) {
    LL X,Y; read(X,Y);
    if(X>Y) swap(X,Y);
    LL t=1,res=0;
    while(F[t+1]<=X&&F[t+2]<=Y) ++t;
    if(t==1) res=X%P*Y%P;
    else {
      for(auto v: G[t-1]) {
        LL x=v.Fst,y=v.Snd;
        if(y<=X) (res+=(Y-x)/y)%=P;
        if(y<=Y) (res+=(X-x)/y)%=P;
      }
    }
    printf("%lld %lld\n",t,res);
  }
	return 0;
}

```