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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=105,p=1000000007;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
ll readl(){
	ll f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int typ,T,k,ans;
ll n,m,f[N];
struct num{
	ll x,y;
	num(ll _x=0,ll _y=0){x=_x;y=_y;}
}a[N][N];
int main(){
//	freopen("gcd.in","r",stdin);
//	freopen("gcd.out","w",stdout);
	f[0]=f[1]=1;
	for (int i=2;i<=90;i++) f[i]=f[i-1]+f[i-2];
	a[0][1]=num(1,1);
	for (int i=1;i<90;i++){
		for (int j=1;j<=i;j++)
		a[i][j]=num(a[i-1][j].x+a[i-1][j].y,a[i-1][j].x);
		a[i][i+1]=num(a[i-1][1].x*2+a[i-1][1].y,a[i-1][1].x);
	}
	typ=1;//read();
	T=read();
	while (T--){
		n=readl();m=readl();
		if (n>m) swap(n,m);
		if (typ==1){
			if (n==1){printf("1 %d\n",m%p);continue;}
			else if (m==2) {puts("1 4");continue;}
			ans=0;
			for (k=1;(f[k+1]<=n)&&(f[k+2]<=m);k++);
			for (int i=1;i<=k;i++){
				ll x=a[k-1][i].x,y=a[k-1][i].y;
				if ((x<=n)&&(y<=m)) ans=(ans+(m-y)/x)%p;
				if ((x<=m)&&(y<=n)) ans=(ans+(n-y)/x)%p;
			}
			printf("%d %d\n",k,ans);
		}
		else{
			ans=((n==1) ? m : (n+m))%p;
			for (k=2;(f[k+1]<=n)&&(f[k+2]<=m);k++)
			for (int i=1;i<=k;i++){
				ll x=a[k-1][i].x,y=a[k-1][i].y;
				if ((x<=n)&&(y<=m)) ans=(ans+(m-y)/x)%p;
				if ((x<=m)&&(y<=n)) ans=(ans+(n-y)/x)%p;
			}
			printf("%d\n",ans);
		}
	}
	return 0;
}