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<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=90;
const int mod=1000000007;
ll f[maxn];
typedef pair<ll,ll> pr;
vector<pr> lv[maxn];
int F(ll a,ll b){
  if(a<b)swap(a,b);
  if(a%b==0)return 1;
  return F(b,a%b)+1;
}
void init(){
  f[0]=1;f[1]=1;
  int tot=1;
  while(f[tot]<=1e18){
    ++tot;f[tot]=f[tot-1]+f[tot-2];
  }
  lv[1].push_back(pr(1,2));lv[1].push_back(pr(1,3));
  ll a,b;
  for(int i=2;i<=tot;++i){
    for(vector<pr>::iterator pt=lv[i-1].begin();pt!=lv[i-1].end();++pt){
      a=pt->first;b=pt->second;
      for(int j=1;j<=2;++j){
	if(b*j+a>f[i+2])break;
	if(F(b,b*j+a)==i){
	  lv[i].push_back(pr(b,b*j+a));
	}
      }
    }
    sort(lv[i].begin(),lv[i].end());
    int N=lv[i].size();
    int cnt=0;
    for(int j=0;j<N;++j){
      if(j==0||lv[i][j]!=lv[i][j-1]){
	lv[i][cnt++]=lv[i][j];
      }
    }//printf("%d\n",cnt);
    lv[i].resize(cnt);
  }
  //  printf("%d %lld\n",tot,f[tot]);
}
void work(ll a,ll b){
  if(a==1||b==1){
    if(a==1){
      printf("%d %d\n",1,b%mod);
    }else{
      printf("%d %d\n",1,a%mod);
    }
    return;
  }
  if(a>b)swap(a,b);
  int t=0;
  while(f[t]<=a&&f[t+1]<=b)++t;
  printf("%d ",t-1);
  if(t-1==1)printf("%d\n",a%mod*1ll*(b%mod)%mod);
  else{
    int ans=0;//printf("%d\n",t-2);
    for(vector<pr>::iterator pt=lv[t-2].begin();pt!=lv[t-2].end();++pt){
      ll x=pt->first,y=pt->second;//printf("%d %d\n",x,y);
      if(x>a||x>b)continue;
      if(y<=a)ans=(ans+(b-x)/y)%mod;
      if(y<=b)ans=(ans+(a-x)/y)%mod;
    }
    printf("%d\n",ans);
  }
}
int main(){
  init();int q;scanf("%d",&q);
  ll a,b;
  while(q--){
    scanf("%lld%lld",&a,&b);
    work(a,b);
  }
  return 0;
}