#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <ctime>
#include <algorithm>
#define N 101
#define PLL pair<ll, ll>
#define MP make_pair
typedef long long ll;
template<typename T> inline void read(T &x) {
  x = 0; char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }
}
using namespace std;
const int P = 1e9 + 7;
int ftot;
ll f[N];
PLL pr[N][N];
inline void work() {
  ll n, m, res = 0; read(n), read(m);
  if (n > m)  swap(n, m);
  if (n == 1) {
    printf("1 %lld\n", m % P);
    return ;
  }
 
  int p;
  p = upper_bound(f + 1, f + 1 + ftot, n) - f - 1;
  if (f[p + 1] > m) --p;
  printf("%d ", p);
  if (p == 1) res = n % P;
  for (int i = 1; i <= p; ++i)
    if (pr[p][i].first <= n && pr[p][i].second <= m)
      res = (res + (m - pr[p][i].second) / pr[p][i].first + 1) % P;
  swap(n, m);
  for (int i = 1; i <= p; ++i)
    if (pr[p][i].first <= n && pr[p][i].second <= m)
      res = (res + (m - pr[p][i].second) / pr[p][i].first + 1) % P;
  printf("%lld\n", (res % P + P) % P);
}
int main() {
  f[0] = f[1] = 1;
  for (int i = 2; f[i - 1] <= 3e18; ++i)  f[i] = f[i - 1] + f[i - 2], ftot = i;
  pr[1][1] = MP(1, 2);
  for (int i = 2; i <= 91; ++i) {
    for (int j = 1; j < i; ++j) pr[i][j] = MP(pr[i - 1][j].second, pr[i - 1][j].first + pr[i - 1][j].second);
    ll a = pr[i - 1][1].first, b = pr[i - 1][1].second;
    pr[i][i] = MP(a + b, a + a + b);
    for (int j = 1; j <= i; ++j) {
    	if (pr[i][j].first > 2e18)	pr[i][j].first = 2e18;
    	if (pr[i][j].second > 2e18)	pr[i][j].second = 2e18;
    }
  }
  int _; read(_);
  while (_--)
    work();
  return 0;
}
### Prompt

Create a solution in CPP for the following 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

### Response

```cpp
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <ctime>
#include <algorithm>
#define N 101
#define PLL pair<ll, ll>
#define MP make_pair
typedef long long ll;
template<typename T> inline void read(T &x) {
  x = 0; char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }
}
using namespace std;
const int P = 1e9 + 7;
int ftot;
ll f[N];
PLL pr[N][N];
inline void work() {
  ll n, m, res = 0; read(n), read(m);
  if (n > m)  swap(n, m);
  if (n == 1) {
    printf("1 %lld\n", m % P);
    return ;
  }
 
  int p;
  p = upper_bound(f + 1, f + 1 + ftot, n) - f - 1;
  if (f[p + 1] > m) --p;
  printf("%d ", p);
  if (p == 1) res = n % P;
  for (int i = 1; i <= p; ++i)
    if (pr[p][i].first <= n && pr[p][i].second <= m)
      res = (res + (m - pr[p][i].second) / pr[p][i].first + 1) % P;
  swap(n, m);
  for (int i = 1; i <= p; ++i)
    if (pr[p][i].first <= n && pr[p][i].second <= m)
      res = (res + (m - pr[p][i].second) / pr[p][i].first + 1) % P;
  printf("%lld\n", (res % P + P) % P);
}
int main() {
  f[0] = f[1] = 1;
  for (int i = 2; f[i - 1] <= 3e18; ++i)  f[i] = f[i - 1] + f[i - 2], ftot = i;
  pr[1][1] = MP(1, 2);
  for (int i = 2; i <= 91; ++i) {
    for (int j = 1; j < i; ++j) pr[i][j] = MP(pr[i - 1][j].second, pr[i - 1][j].first + pr[i - 1][j].second);
    ll a = pr[i - 1][1].first, b = pr[i - 1][1].second;
    pr[i][i] = MP(a + b, a + a + b);
    for (int j = 1; j <= i; ++j) {
    	if (pr[i][j].first > 2e18)	pr[i][j].first = 2e18;
    	if (pr[i][j].second > 2e18)	pr[i][j].second = 2e18;
    }
  }
  int _; read(_);
  while (_--)
    work();
  return 0;
}
```