#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

template <typename T>
inline void read(T &x) {
  x = 0; T f = 1; char c = getchar();
  for(; !isdigit(c); c = getchar())
    if(c == '-') f = -1;
  for(;  isdigit(c); c = getchar())
    x = x * 10 + c - '0';
  x *= f;
}

# define pi pair<LL, LL>
# define pb push_back

namespace pre {
  const int N = 100;

  int tot, up;
  LL fib[N];
  pi g[N];
  vector<pi> exce[N];
  
  inline int f(LL a, LL b) {
    if(!a) return 0;
    if(a > b) return f(b, a);
    return f(b % a, a) + 1;
  }
  void dfs(int d, LL x, LL y) {
    if(x > fib[d + 2] || y > fib[d + 2]) return ;
    if(d == up) {
      exce[d].pb(pi(x, y));
      return ;
    }
    dfs(d + 1, y, x + y);
    dfs(d + 1, y, x + 2ll * y);
  }

  void main() {
    fib[0] = fib[1] = 1;
    for(tot = 1; fib[tot] + fib[tot - 1] <= 6e18; ++tot)
      fib[tot + 1] = fib[tot] + fib[tot - 1];
    while(fib[tot] > 1e18) --tot;
    for(int i = 1; i <= tot; ++i) 
      up = i, dfs(0, 1, 1);
  }
}
using namespace pre;

const int mod = 1e9 + 7;

int cas;
LL n, m;

int main() {
  pre :: main();
  for(cin >> cas; cas--; ) {
    read(n), read(m);
    if(n > m) swap(n, m);
    int get = -1;
    for(int i = tot; ~i; --i) 
      if(fib[i] <= n && fib[i + 1] <= m) {
        get = i;
        break;
      }        
    assert(get >= 0);
    if(get <= 1) {
      printf("1 %lld\n", n == 1 ? m % mod : 4ll);
      continue;
    }
    LL ans = 0;
    for(auto p : exce[get - 1]) {
      LL x = p.first;
      LL y = p.second;
      if(y <= n && x + y <= m) ans += (m - x) / y;
      if(y <= m && x + y <= n) ans += (n - x) / y;
    }
    printf("%d %lld\n", get, ans % mod);
  }
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <bits/stdc++.h>
using namespace std;
typedef long long LL;

template <typename T>
inline void read(T &x) {
  x = 0; T f = 1; char c = getchar();
  for(; !isdigit(c); c = getchar())
    if(c == '-') f = -1;
  for(;  isdigit(c); c = getchar())
    x = x * 10 + c - '0';
  x *= f;
}

# define pi pair<LL, LL>
# define pb push_back

namespace pre {
  const int N = 100;

  int tot, up;
  LL fib[N];
  pi g[N];
  vector<pi> exce[N];
  
  inline int f(LL a, LL b) {
    if(!a) return 0;
    if(a > b) return f(b, a);
    return f(b % a, a) + 1;
  }
  void dfs(int d, LL x, LL y) {
    if(x > fib[d + 2] || y > fib[d + 2]) return ;
    if(d == up) {
      exce[d].pb(pi(x, y));
      return ;
    }
    dfs(d + 1, y, x + y);
    dfs(d + 1, y, x + 2ll * y);
  }

  void main() {
    fib[0] = fib[1] = 1;
    for(tot = 1; fib[tot] + fib[tot - 1] <= 6e18; ++tot)
      fib[tot + 1] = fib[tot] + fib[tot - 1];
    while(fib[tot] > 1e18) --tot;
    for(int i = 1; i <= tot; ++i) 
      up = i, dfs(0, 1, 1);
  }
}
using namespace pre;

const int mod = 1e9 + 7;

int cas;
LL n, m;

int main() {
  pre :: main();
  for(cin >> cas; cas--; ) {
    read(n), read(m);
    if(n > m) swap(n, m);
    int get = -1;
    for(int i = tot; ~i; --i) 
      if(fib[i] <= n && fib[i + 1] <= m) {
        get = i;
        break;
      }        
    assert(get >= 0);
    if(get <= 1) {
      printf("1 %lld\n", n == 1 ? m % mod : 4ll);
      continue;
    }
    LL ans = 0;
    for(auto p : exce[get - 1]) {
      LL x = p.first;
      LL y = p.second;
      if(y <= n && x + y <= m) ans += (m - x) / y;
      if(y <= m && x + y <= n) ans += (n - x) / y;
    }
    printf("%d %lld\n", get, ans % mod);
  }
}
```