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 <algorithm>
#include <cstdio>

typedef long long LL;
const int mo=1000000007;
LL f[100],a[100][200],b[100][200],cnt[100];

int main(){
	f[0]=f[1]=1;
	for (int i=2;i<90;i++) f[i]=f[i-1]+f[i-2];
	cnt[1]=2;a[1][0]=1;b[1][0]=2;a[1][1]=1;b[1][1]=3;
	for (int i=1;i<85;i++)
		for (int j=0;j<cnt[i];j++){
			LL x=b[i][j],y=a[i][j]+b[i][j];
			for (;y<=f[i]+f[i+3];y+=b[i][j])
				a[i+1][cnt[i+1]]=x,b[i+1][cnt[i+1]++]=y;
		}
	int Q;
	scanf("%d\n",&Q);
	for (;Q--;){
		LL X,Y;int ans=0;
		scanf("%lld%lld\n",&X,&Y);
		if (X>Y) std::swap(X,Y);
		int k=1;
		for (;f[k+1]<=X && f[k+2]<=Y;k++);
		if (!--k) ans=1ll*X*Y%mo;
		else{
			for (int i=0;i<cnt[k];i++){
				LL x=a[k][i],y=b[k][i];
				if (y<=X) ans=(ans+(Y-x)/y)%mo;
				if (y<=Y && x<=X) ans=(ans+(X-x)/y)%mo;
			}
		}
		printf("%d %d\n",k+1,ans);
	}
}