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;
const int q=1e9+7;
typedef long long giant;
typedef pair<giant,giant> Pair;
typedef pair<int,int> Data;
namespace num {
	const int maxn=91;
	giant F[maxn];
	vector<Pair> a[maxn];
	void init() {
		F[0]=F[1]=1;
		for (int i=2;i<maxn;++i) F[i]=F[i-1]+F[i-2];
		assert(F[maxn-1]>2e18);
		a[1].emplace_back(1,2);
		a[1].emplace_back(1,3);
		for (int i=2;i+2<maxn;++i) {
			for (const Pair &pr:a[i-1]) {
				Pair nw(pr.second,pr.first+pr.second);
				giant lim=F[i-1]+F[i+2];
				for (;nw.second<lim;nw.second+=nw.first) a[i].push_back(nw);
			}
		}
	}
	int __initer__=(init(),0);
}
int T,ans;
giant A,B,cnt;
Data work() {
	scanf("%lld%lld",&A,&B);
	if (A>B) swap(A,B);
	ans=1;
	for (int i=1;num::F[i]<=A && num::F[i+1]<=B;++i) ans=i;
	if (ans==1) return Data(1,A%q*B%q);
	cnt=0;
	for (const Pair &pr:num::a[ans-1]) {
		if (pr.second<=A) cnt=(cnt+(B-pr.first)/pr.second)%q;
		if (pr.second<=B) cnt=(cnt+(A-pr.first)/pr.second)%q;
	}
	return Data(ans,cnt);
}
int main() {
	for (scanf("%d",&T);T--;) {
		Data ans(work());
		printf("%lld %lld\n",ans.first,ans.second);
	}
	return 0;
}