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;
#define LL long long
const int mod=1e9+7;
const LL maxN=4e18;
LL fib[205];
LL getnum(LL x,LL y,LL X,LL Y){//x<y
    LL ret=0;
    if(x<=X&&y<=Y) ret+=(Y-y+x)/x;
    if(x<=Y&&y<=X) ret+=(X-y+x)/x;
    return ret%mod;
}
void solve(LL X,LL Y){
    if(X>Y) swap(X,Y);
    int k=1;
    while(fib[k+1]<=X&&fib[k+2]<=Y) k++;
    printf("%d ",k);
    if(k==1) {printf("%lld\n",X*Y%mod);return;}
    LL ret=getnum(fib[k],fib[k+1],X,Y);
    for(int i=1;i<k;i++)
        ret+=getnum(fib[i]*fib[k-1-i] +fib[k] ,fib[i]*fib[k-i] +fib[k+1] ,X,Y);
    printf("%lld\n",ret%mod);
}
void pre(){
    fib[0]=fib[1]=1;
    for(int i=1;fib[i]<=maxN;i++) fib[i+1]=fib[i]+fib[i-1];
}

int main(){
    pre();
    int T;
    scanf("%d",&T);
    while(T--){
        LL X,Y;
        scanf("%lld%lld",&X,&Y);
        solve(X,Y);
    }
    return 0;
}