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

Solution (in CPP):

Solution:
#include<cstdio>
typedef long long ll;
const int mod=1000000007;
const ll MAX=1000000000000000000ll;
ll f[100];
int lim;
struct Point{ll x,y;}a[100][100];
int calc(ll X,ll Y){
	if(X>Y)return calc(Y,X);
	int i=1;
	while(i+1<lim&&X>=f[i+1]&&Y>=f[i+2])i++;
	printf("%d ",i);
	if(i==1)return X%mod*(Y%mod)%mod;
	ll s=0;
	for(Point*p=a[i];p<=a[i]+i;p++)if(p->x<=X&&p->y<2*p->x){
		if(p->y<=Y)s+=(Y-p->y)/p->x+1;
		if(p->y<=X)s+=(X-p->y)/p->x+1;
		s%=mod;
	}
	return s;
}
int main(){
	lim=1,f[0]=f[1]=1;
	for(;f[lim-1]+f[lim]<=MAX;lim++)f[lim+1]=f[lim-1]+f[lim];
	a[1][0]=(Point){1,2};
	a[1][1]=(Point){1,3};
	for(int i=2;i<lim;i++){
		for(Point*j=a[i-1],*k=a[i];j<a[i-1]+i;j++){
			*k++=(Point){j->y,j->x+j->y};
			if(j->x==f[i-1]&&j->y==f[i])*k++=(Point){j->y,j->x+j->y*2};
		}
	}
	int Q;scanf("%d",&Q);
	while(Q--){
		ll X,Y;
		scanf("%lld%lld",&X,&Y);
		printf("%d\n",calc(X,Y));
	}
}