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:
//神题
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define mod 1000000007
using namespace std;
struct ww{ll a, b;}w[90][90];
int n, ans;
ll f[90], x, y, sum;
int main(){
	f[0]=f[1]=1; for(int i=2; i<=88; i++)f[i]=f[i-2]+f[i-1];
	w[1][1].a=1; w[1][1].b=2; w[1][2].a=1; w[1][2].b=3;
	for(int i=1; i<=86; i++){
		for(int j=1; j<=i+1; j++){w[i+1][j].a=w[i][j].b; w[i+1][j].b=w[i][j].a+w[i][j].b;}
		w[i+1][i+2].a=f[i+1]; w[i+1][i+2].b=f[i+3];
	}
	scanf("%d", &n);
	for(int owo=1; owo<=n; owo++){
		scanf("%lld%lld", &x, &y); if(x>y)swap(x, y);
		ans=1; for(int i=2; i<=86; i++)if(f[i]<=x&&f[i+1]<=y)ans=i;
		printf("%d ", ans);
		if(ans==1){printf("%d\n", (x%mod)*(y%mod)%mod); continue;}
		sum=0; ans--;
		for(int i=1; i<=ans+1; i++){
			sum=(sum+(x-w[ans][i].a)/w[ans][i].b)%mod;
			if(x>=w[ans][i].b)sum=(sum+(y-w[ans][i].a)/w[ans][i].b)%mod;
		}
		printf("%d\n", sum);
	}
	return 0;
}