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<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>

using namespace std;

typedef long long LL;
const int MX=201;
const LL mxx=1000000000000000000LL,md=1000000007;
//int f[MX][MX];
typedef pair<LL,LL> pii;
#define mp make_pair
#define se second
#define fr first
vector<pii>S[MX];
typedef vector<pii>::iterator vit;

int main(){
	S[1].push_back(mp(1,2));
	int it=1;
	while(S[it][0].fr<=mxx){
		++it;
		for(vit i=S[it-1].begin();i!=S[it-1].end();i++)
			S[it].push_back(mp(i->se,i->fr+i->se));
		S[it].push_back(mp(S[it-1][0].fr+S[it-1][0].se,S[it-1][0].fr*2+S[it-1][0].se));
	}
	int q;scanf("%d",&q);
	while(q--){
		LL x,y;scanf("%lld%lld",&x,&y);
		if(x>y)swap(x,y);
		if(x==1)printf("%d %lld\n",1,y%md);
		else if(x==2&&y==2)puts("1 4");
		else{
			int it=1;while(S[it][0].fr<=x&&S[it][0].se<=y)++it;
			--it;
			LL ans=0LL;
			for(vit i=S[it].begin();i!=S[it].end();i++){
				if(x>=i->fr&&y>=i->se)ans=(ans+(y-i->se)/i->fr+1)%md;
				if(x>=i->se&&y>=i->fr)ans=(ans+(x-i->se)/i->fr+1)%md;
			}
			printf("%d %lld\n",it,ans);
		}
	}
	return 0;
}