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>
#define ll long long
#define mo 1000000007
#define pll pair<ll,ll>
using namespace std;
ll f[105];
vector<pll> s[105];
void dfs(int x,ll a,ll b){
	if (x>88) return;
	if (a>=f[x]&&b>=f[x+1]) return;
	s[x].push_back(pll(a,b));
	dfs(x+1,b,b+a); dfs(x+1,b,b*2+a);
}
int main(){
	f[0]=f[1]=1;
	for (int i=2;i<=95;i++)
		f[i]=f[i-1]+f[i-2];
	dfs(1,1,1);
	int Q;
	scanf("%d",&Q);
	while (Q--){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		if (x>y) swap(x,y);
		int ans=0; ll sum=0;
		for (int i=1;i<=100;i++)
			if (f[i]>x||f[i+1]>y){
				ans=i-1;
				break;
			}
		if (ans<=1){
			if (x==1) sum=y%mo;
			else if (x==2) sum=4;
			printf("1 %lld\n",sum);
			continue;
		}
		for (int i=0;i<s[ans].size();i++){
			pll p=s[ans][i];
			if (p.second<=x&&y>=p.first)
				sum=(sum+(y-p.first)/p.second)%mo;
			if (p.second<=y&&x>=p.first)
				sum=(sum+(x-p.first)/p.second)%mo;
		}
		printf("%d %lld\n",ans,sum);
	}
}