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<algorithm>
#define ll long long
#define MN 500
using namespace std;
ll read_p,read_ca,read_f;
inline ll read(){
	read_p=0;read_ca=getchar();read_f=1;
	while (read_ca<'0'||read_ca>'9') read_f=read_ca=='-'?-1:read_f,read_ca=getchar();
	while (read_ca>='0'&&read_ca<='9') read_p=read_p*10+read_ca-48,read_ca=getchar();
	return read_p*read_f;
}
int T;
const ll M=1e18;
const int MOD=1e9+7;
struct na{ll x,y;}S[MN][MN];
ll n,m,F[MN],mmh,MMH;
int main(){
	T=read();
	F[0]=F[1]=1;
	for (ll i=2;(F[i]=F[i-1]+F[i-2])<=M;i++);
	S[0][0].x=S[0][0].y=1;
	for (ll i=1;F[i]<=M;i++){
		for (ll j=0;j<i;j++) S[i][j].x=S[i-1][j].x+S[i-1][j].y,S[i][j].y=S[i-1][j].x;
		S[i][i].x=F[i+2];S[i][i].y=F[i];
	}
	while(T--){
		n=read();m=read();
		if (n==1&&m==1) {printf("1 1\n");continue;}
		if (n<m) swap(n,m);
		mmh=1;
		while (F[mmh+2]<=n&&F[mmh+1]<=m) mmh++;
		printf("%lld ",mmh);mmh--;
		MMH=0;
		for (ll i=0;i<=mmh;i++){
			if (m>=S[mmh][i].x) (MMH+=(n-S[mmh][i].y)/S[mmh][i].x)%=MOD;
			if (n>=S[mmh][i].x) (MMH+=(m-S[mmh][i].y)/S[mmh][i].x)%=MOD;
		}
		printf("%lld\n",(MMH+(mmh?0:m))%MOD);
	}
}