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>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1e9+7;
typedef long long ll;
int n,mx,tot,head[110],nxt[500010];
ll f[110],valx[500010],valy[500010];
void add(int u,ll x,ll y){
	nxt[++tot]=head[u];
	valx[tot]=x;
	valy[tot]=y;
	head[u]=tot;
	return;
}
void init(){
	memset(head,-1,sizeof(head));
	f[0]=1;
	f[1]=1;
	for(mx=2;f[mx-1]<=1e18;mx++)
		f[mx]=f[mx-1]+f[mx-2];
	mx--;
	add(1,1,2);
	add(1,1,3);
	add(1,1,4);
	for(int i=2;i<=mx-2;i++){
		for(int j=head[i-1];~j;j=nxt[j]){
			ll x=valx[j],y=valy[j];
			for(x+=y;x<=f[i+2];x+=y)
				add(i,y,x);
		}
	}
	return;
}
void solve(ll X,ll Y){
	if(X>Y)
		swap(X,Y);
	int d,ans=0;
	for(d=1;f[d+1]<=X&&f[d+2]<=Y;d++);
	printf("%d ",d);
	d--;
	if(!d){
		printf("%d\n",X*Y%mod);
		return;
	}
	for(int i=head[d];~i;i=nxt[i]){
		ll x=valx[i],y=valy[i];
		if(X>=y)
			ans=(ans+(Y-x)/y)%mod;
		if(Y>=y)
			ans=(ans+(X-x)/y)%mod;
	}
	printf("%d\n",ans);
}
int main(){
	init();
	scanf("%d",&n);
	while(n--){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		solve(x,y);
	}
	return 0;
}