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:
//agc015f
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<time.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define fe(x) for(int ii=be[x];ii;ii=e[ii].ne)
using namespace std;
const int ft=90,p=1e9+7;
int q,len;
long long n,m,f[100],ans;
struct nod{
	long long a,b;
};
nod g[100][100];
int main(){
	f[1]=1;f[2]=2;fo(i,3,ft) f[i]=f[i-1]+f[i-2];
	g[0][0].a=g[0][0].b=1;
	fo(i,1,ft-1){
		fo(j,0,i-1){
			g[i][j].a=g[i-1][j].a+g[i-1][j].b;
			g[i][j].b=g[i-1][j].a;
		}
		g[i][i].a=f[i+2];g[i][i].b=f[i];
		/*if (i==ft-1){
			printf("------%d----\n",i);
			fo(j,0,i) printf("%I64d %I64d\n",g[i][j].a,g[i][j].b);
		}*/
	}
	scanf("%d",&q);
	while (q--){
		scanf("%lld%lld",&n,&m);
		if (n>m) swap(n,m);
		if (n==1){
			printf("1 %lld\n",m%p);
			continue;
		}
		fd(i,ft-1,1) if (f[i]<=n&&f[i+1]<=m){
			len=i;
			break;
		}
		if (len==1){
			printf("1 %lld\n",(n%p)*(m%p)%p);
			continue;
		}
		ans=0;
		fo(i,0,len-1){//0..len!!!!!
			if (g[len][i].b<=n&&g[len][i].a<=m) ans+=(m-g[len][i].a)/g[len][i].b+1;
			if (g[len][i].a<=n) ans++;
			ans%=p;
		}
		printf("%d %lld\n",len,ans);
	}
	return 0;
}