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>

typedef long long i64;

const int N = 90, MOD = 1000000007;

i64 n, m, fib[N];

std::vector<std::pair<i64, i64>> pool[N];

int main() {
  fib[0] = fib[1] = 1;
  for (int i = 2; i < N; ++i) fib[i] = fib[i - 1] + fib[i - 2];
  pool[1].emplace_back(1, 2);
  pool[1].emplace_back(1, 3);
  for (int i = 2; i + 2 < N; ++i)
    for (auto it : pool[i - 1])
      for (i64 y = it.first + it.second; y <= fib[i + 2]; y += it.second)
        pool[i].emplace_back(it.second, y);
  int q;
  for (scanf("%d", &q); q--;) {
    scanf("%lld%lld", &n, &m);
    if (n > m) std::swap(n, m);
    if (n == 1) {
      printf("1 %d\n", m % MOD);
    } else if (n == 2 && m == 2) {
      puts("1 4");
    } else {
      int k = 1;
      while (fib[k + 1] <= n && fib[k + 2] <= m) ++k;
      int ans = 0;
      for (auto it : pool[k - 1]) {
        if (it.second <= n) ans = (ans + (m - it.first) / it.second) % MOD;
        if (it.second <= m) ans = (ans + (n - it.first) / it.second) % MOD;
      }
      printf("%d %d\n", k, ans);
    }
  }
  return 0;
}