#include <cstdio>
#include <cstring>
#include <cctype>
#include <vector>
using namespace std;
typedef long long int64;
inline int64 read(int f = 1, int64 x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 90+5, P = 1e9+7;
int n, m, ans, q;
int64 f[N];
vector<pair<int64, int64> > s[N];
int main()
{
    for(f[0] = f[1] = 1, n = 2; n <= 90; ++n) f[n] = f[n-1]+f[n-2]; --n;
    s[1].push_back(make_pair(1, 2)), s[1].push_back(make_pair(1, 3)), s[1].push_back(make_pair(1, 4));
    for(int i = 1; i <= n; ++i)
        for(int j = 0; j < s[i].size(); ++j)
        {
            int64 x = s[i][j].second, y = s[i][j].first+x;
            while(y <= f[i]+f[i+3]) s[i+1].push_back(make_pair(x, y)), y += x;
        }
    q = read();
    while(q--)
    {   
        int64 x = read(), y = read(); if(x > y) swap(x, y);
        ans = 0; for(m = 1; f[m+1] <= x&&f[m+2] <= y; ++m);
        if(m == 1) ans = x*y%P;
        else
            for(int i = 0; i < s[m-1].size(); ++i)
            {
                int64 u = s[m-1][i].first, v = s[m-1][i].second;
                if(v <= x) ans = (ans+(y-u)/v)%P;
                if(v <= y) ans = (ans+(x-u)/v)%P;
            }
        printf("%d %d\n", m, ans);
    }
    return 0;
}
### Prompt

Please formulate a Cpp solution 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 <cstdio>
#include <cstring>
#include <cctype>
#include <vector>
using namespace std;
typedef long long int64;
inline int64 read(int f = 1, int64 x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 90+5, P = 1e9+7;
int n, m, ans, q;
int64 f[N];
vector<pair<int64, int64> > s[N];
int main()
{
    for(f[0] = f[1] = 1, n = 2; n <= 90; ++n) f[n] = f[n-1]+f[n-2]; --n;
    s[1].push_back(make_pair(1, 2)), s[1].push_back(make_pair(1, 3)), s[1].push_back(make_pair(1, 4));
    for(int i = 1; i <= n; ++i)
        for(int j = 0; j < s[i].size(); ++j)
        {
            int64 x = s[i][j].second, y = s[i][j].first+x;
            while(y <= f[i]+f[i+3]) s[i+1].push_back(make_pair(x, y)), y += x;
        }
    q = read();
    while(q--)
    {   
        int64 x = read(), y = read(); if(x > y) swap(x, y);
        ans = 0; for(m = 1; f[m+1] <= x&&f[m+2] <= y; ++m);
        if(m == 1) ans = x*y%P;
        else
            for(int i = 0; i < s[m-1].size(); ++i)
            {
                int64 u = s[m-1][i].first, v = s[m-1][i].second;
                if(v <= x) ans = (ans+(y-u)/v)%P;
                if(v <= y) ans = (ans+(x-u)/v)%P;
            }
        printf("%d %d\n", m, ans);
    }
    return 0;
}
```