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;
typedef long long ll;
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
const int M=1e9+7;
vector<pii > V[300];
ll f[300],n,m;
int T;
int main(){
	V[1].push_back(mp(1,2));
	V[1].push_back(mp(1,3));
	V[1].push_back(mp(1,4));
	f[0]=f[1]=1;
	for (int i=2;i<300;i++)f[i]=f[i-1]+f[i-2];
	for (int i=1;i;i++){
		if (f[i+1]>1e18)break;
		for (pii j:V[i]){
			ll x=j.fi,y=j.fi;
			while (y+j.se<=f[i+3]+f[i]){
				y+=j.se;
				V[i+1].push_back(mp(j.se,y));
			}
		}
	}
	scanf("%d",&T);
	while (T--){
		scanf("%lld%lld",&n,&m);
		if (n>m)swap(n,m);
		int ans=1;
		for (;f[ans+1]<=n&&f[ans+2]<=m;ans++);
		printf("%d ",ans);
		if (ans==1){
			printf("%lld\n",n*m%M);
			continue;
		}
		ll ans2=0;
		for (pii j:V[ans-1]){
			if (j.se<=n&&j.se<=m)(ans2+=(m-j.fi)/j.se)%=M;
			if (j.se<=m&&j.fi<=n)(ans2+=(n-j.fi)/j.se)%=M;
		}
		printf("%lld\n",ans2);
	}
}