# include <vector>
# include <stdio.h>
# include <string.h>
# include <iostream>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

const int M = 5e5 + 10, N = 95;
const int mod = 1e9 + 7;

vector < pair <ll, ll> > g[N];
ll f[N];


int main() {
//  freopen("euclid.in", "r", stdin);
//  freopen("euclid.out", "w", stdout);
  ios :: sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  f[0] = f[1] = 1;
  for (int i = 2; i < N; ++ i) f[i] = f[i-1] + f[i-2];
  g[1].push_back(make_pair(1, 2));
  g[1].push_back(make_pair(1, 3));
  g[1].push_back(make_pair(1, 4));
  for (int i = 1; i < N; ++ i) {
    for (int j = 0; j < g[i].size(); ++ j) {
      pair <ll, ll> cur = g[i][j];
      ll x = cur.second, y = cur.first + cur.second;
      while(y <= f[i+3] + f[i]) {
        g[i+1].push_back(make_pair(x, y));
        y += x;
      }
    }
  }
  
  int q; cin >> q;
  ll a, b;
  while(q--) {
    cin >> a >> b;
    if(a > b) swap(a, b);
    int ans = 1;
    while(f[ans+1] <= a && f[ans+2] <= b) ++ ans;
    cout << ans << ' ';
    if(ans == 1) cout << 1ll * a * b % mod << '\n';
    else {
      int cnt = 0;
      for (int i = 0; i < g[ans-1].size(); ++ i) {
        pair<ll, ll> cur = g[ans-1][i];
        ll x = cur.first, y = cur.second;
        if(y <= a) cnt = cnt + 1ll * (b - x) / y % mod;
        if(cnt >= mod) cnt -= mod;
        if(y <= b) cnt = cnt + 1ll * (a - x) / y % mod;
        if(cnt >= mod) cnt -= mod; 
      }
      cout << cnt << '\n';
    }
  }
  
  
  
  return 0;
}



### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
# include <vector>
# include <stdio.h>
# include <string.h>
# include <iostream>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

const int M = 5e5 + 10, N = 95;
const int mod = 1e9 + 7;

vector < pair <ll, ll> > g[N];
ll f[N];


int main() {
//  freopen("euclid.in", "r", stdin);
//  freopen("euclid.out", "w", stdout);
  ios :: sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  f[0] = f[1] = 1;
  for (int i = 2; i < N; ++ i) f[i] = f[i-1] + f[i-2];
  g[1].push_back(make_pair(1, 2));
  g[1].push_back(make_pair(1, 3));
  g[1].push_back(make_pair(1, 4));
  for (int i = 1; i < N; ++ i) {
    for (int j = 0; j < g[i].size(); ++ j) {
      pair <ll, ll> cur = g[i][j];
      ll x = cur.second, y = cur.first + cur.second;
      while(y <= f[i+3] + f[i]) {
        g[i+1].push_back(make_pair(x, y));
        y += x;
      }
    }
  }
  
  int q; cin >> q;
  ll a, b;
  while(q--) {
    cin >> a >> b;
    if(a > b) swap(a, b);
    int ans = 1;
    while(f[ans+1] <= a && f[ans+2] <= b) ++ ans;
    cout << ans << ' ';
    if(ans == 1) cout << 1ll * a * b % mod << '\n';
    else {
      int cnt = 0;
      for (int i = 0; i < g[ans-1].size(); ++ i) {
        pair<ll, ll> cur = g[ans-1][i];
        ll x = cur.first, y = cur.second;
        if(y <= a) cnt = cnt + 1ll * (b - x) / y % mod;
        if(cnt >= mod) cnt -= mod;
        if(y <= b) cnt = cnt + 1ll * (a - x) / y % mod;
        if(cnt >= mod) cnt -= mod; 
      }
      cout << cnt << '\n';
    }
  }
  
  
  
  return 0;
}



```