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>
typedef long long i64;
const int P=1000000007;
int T;
i64 a,b,f[107],fs[107][107][2];
int main(){
	f[0]=f[1]=1;
	for(int i=2;i<=100;++i)f[i]=f[i-1]+f[i-2];
	fs[0][0][0]=fs[0][0][1]=1;
	for(int i=1;i<=100;++i){
		for(int j=0;j<i;++j){
			fs[i][j][0]=fs[i-1][j][0]+fs[i-1][j][1];
			fs[i][j][1]=fs[i-1][j][0];
		}
		fs[i][i][0]=f[i+2];fs[i][i][1]=f[i];
	}
	for(scanf("%d",&T);T;--T){
		scanf("%lld%lld",&a,&b);
		if(a>b)std::swap(a,b);
		int mv=1;
		for(;f[mv+1]<=a&&f[mv+2]<=b;++mv);
		if(mv==1)printf("1 %lld\n",a*b%P);
		else{
			i64 s=0;
			for(int i=0;i<mv;++i){
				i64 x=fs[mv-1][i][0],y=fs[mv-1][i][1];
				if(b>=x)s+=(a-y)/x;
				if(a>=x)s+=(b-y)/x;
				if(s>6e18||s<-6e18)s%=P;
			}
			printf("%d %lld\n",mv,(s%P+P)%P);
		}
	}
	return 0;
}