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 <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int P=1e9+7,N=90;
typedef long long ll;
int T;ll n,m,f[N+10];
vector<pair<ll,ll> > g[N+10];
int main(){
    scanf("%d",&T);
    f[0]=1;f[1]=1;
    rep(i,2,N) f[i]=f[i-1]+f[i-2];
    g[1]={mp(1,2),mp(1,3),mp(1,4)};
    rep(i,1,N-1) for(auto p:g[i]){
        ll x=p.se,y=p.fi+p.se;
        while(y<=f[i+3]+f[i]) g[i+1].emplace_back(x,y),y+=x;
    }
    while(T--){
        scanf("%lld%lld",&n,&m);if(n>m) swap(n,m);
        int t=1;
        while(f[t+1]<=n&&f[t+2]<=m) ++t;
        printf("%d ",t);
        if(t==1){printf("%lld\n",n%P*(m%P)%P);continue;}--t;
        ll ans=0;
        for(auto p:g[t]){
            if(p.se<=n) ans+=(m-p.fi)/p.se%P;
            if(p.se<=m) ans+=(n-p.fi)/p.se%P;
        }
        printf("%lld\n",ans%P);
    }
    return 0;
}