#include <bits/stdc++.h>
#define RI register int
typedef long long LL;

#define FILEIO(name) freopen(name".in", "r", stdin), freopen(name".out", "w", stdout);

using namespace std;

char buf[1000000], *p1 = buf, *p2 = buf;
inline char gc() {
  if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);
  return p1 == p2 ? EOF : *(p1++);
}
template <class T> inline void read(T &n) {
  n = 0; RI ch = gc(), f;
  while ((ch < '0' || ch > '9') && ch != '-') ch = gc();
  f = (ch == '-' ? ch = gc(), -1 : 1);
  while (ch >= '0' && ch <= '9') n = n * 10 + (ch ^ 48), ch = gc();
  n *= f;
}

int const mod = 1e9 + 7;
LL f[105], x, y;
int lim = 91;
LL ans2;

bool Dfs(LL a, LL b, LL res) {
  if (a * f[res - 1] + (res - 1 ? b * f[res - 2] : 0) > x) return 0;
  if (a * f[res] + b * f[res - 1] > y) return 0;
  if (!(res - 1)) {
    ans2 = (ans2 + (x - b) / a) % mod;
    ans2 = (ans2 + (y - b) / a) % mod;
    return 1;
  }
  int i = 1;
  while (Dfs(b + i * a, a, res - 1)) ++i;
  return (i > 1);
}

int main() {
  
#ifdef LOCAL
  FILEIO("a");
#endif

  f[0] = f[1] = 1;
  for (RI i = 2; i <= lim; ++i)
    f[i] = f[i - 1] + f[i - 2];
  int Q; read(Q);
  while (Q--) {
    read(x), read(y);
    if (x > y) swap(x, y);
    if (x == 1) { printf("%d %lld\n", 1, y % mod); continue; }
    int ans1 = 0;
    for (RI i = lim; i && !ans1; --i)
      if (f[i - 1] <= x && f[i] <= y)
        ans1 = i - 1;
    ans2 = 0;
    if (ans1 == 1) {
      ans2 = (ans2 + x) % mod;
      ans2 = (ans2 + y) % mod;
    }
    else for (RI i = 2; Dfs(i, 1, ans1 - 1); ++i);
    printf("%d %lld\n", ans1, ans2);
  }

  return 0;
}

// created by Daniel yuan

### Prompt

Please create a solution in cpp to 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 <bits/stdc++.h>
#define RI register int
typedef long long LL;

#define FILEIO(name) freopen(name".in", "r", stdin), freopen(name".out", "w", stdout);

using namespace std;

char buf[1000000], *p1 = buf, *p2 = buf;
inline char gc() {
  if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);
  return p1 == p2 ? EOF : *(p1++);
}
template <class T> inline void read(T &n) {
  n = 0; RI ch = gc(), f;
  while ((ch < '0' || ch > '9') && ch != '-') ch = gc();
  f = (ch == '-' ? ch = gc(), -1 : 1);
  while (ch >= '0' && ch <= '9') n = n * 10 + (ch ^ 48), ch = gc();
  n *= f;
}

int const mod = 1e9 + 7;
LL f[105], x, y;
int lim = 91;
LL ans2;

bool Dfs(LL a, LL b, LL res) {
  if (a * f[res - 1] + (res - 1 ? b * f[res - 2] : 0) > x) return 0;
  if (a * f[res] + b * f[res - 1] > y) return 0;
  if (!(res - 1)) {
    ans2 = (ans2 + (x - b) / a) % mod;
    ans2 = (ans2 + (y - b) / a) % mod;
    return 1;
  }
  int i = 1;
  while (Dfs(b + i * a, a, res - 1)) ++i;
  return (i > 1);
}

int main() {
  
#ifdef LOCAL
  FILEIO("a");
#endif

  f[0] = f[1] = 1;
  for (RI i = 2; i <= lim; ++i)
    f[i] = f[i - 1] + f[i - 2];
  int Q; read(Q);
  while (Q--) {
    read(x), read(y);
    if (x > y) swap(x, y);
    if (x == 1) { printf("%d %lld\n", 1, y % mod); continue; }
    int ans1 = 0;
    for (RI i = lim; i && !ans1; --i)
      if (f[i - 1] <= x && f[i] <= y)
        ans1 = i - 1;
    ans2 = 0;
    if (ans1 == 1) {
      ans2 = (ans2 + x) % mod;
      ans2 = (ans2 + y) % mod;
    }
    else for (RI i = 2; Dfs(i, 1, ans1 - 1); ++i);
    printf("%d %lld\n", ans1, ans2);
  }

  return 0;
}

// created by Daniel yuan

```