#include<bits/stdc++.h>
using namespace std;
#define MAXN	105
#define INF	1e18
#define P	1000000007
long long fib[MAXN];
vector<pair<long long, long long> > a[MAXN];
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
int main(){
	fib[0] = fib[1] = 1;
	int tot = 1;
	while (true) {
		fib[tot + 1] = fib[tot] + fib[tot - 1];
		if (fib[tot + 1] > INF) break;
		tot++;
	}
	a[1].push_back(make_pair(1, 2));
	a[1].push_back(make_pair(1, 3));
	a[1].push_back(make_pair(1, 4));
	for (int i = 1; i <= tot - 3; i++) {
		for (unsigned j = 0; j < a[i].size(); j++) {
			long long x = a[i][j].first;
			long long y = a[i][j].second;
			x += y;
			while (x <= fib[i + 3] + fib[i]){
				a[i + 1].push_back(make_pair(y, x));
				x += y;
			}
		}
	}
	int q; read(q);
	while (q--){
		long long n, m;
		read(n), read(m);
		if (n > m) swap(n, m);
		int ans = 1;
		while (fib[ans + 1] <= n && fib[ans + 2] <= m)
			ans++;
		printf("%d ", ans);
		if (ans == 1) {
			printf("%lld\n", n * m % P);
			continue;
		}
		else {
			long long num = 0;
			for (unsigned j = 0; j < a[ans - 1].size(); j++) {
				long long x = a[ans - 1][j].first;
				long long y = a[ans - 1][j].second;
				if (x <= m && y <= n) num = (num + (m - x) / y) % P;
				if (x <= n && y <= m) num = (num + (n - x) / y) % P;
			}
			printf("%lld\n", num);
		}
	}
}
### Prompt

Your task is to create 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<bits/stdc++.h>
using namespace std;
#define MAXN	105
#define INF	1e18
#define P	1000000007
long long fib[MAXN];
vector<pair<long long, long long> > a[MAXN];
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
int main(){
	fib[0] = fib[1] = 1;
	int tot = 1;
	while (true) {
		fib[tot + 1] = fib[tot] + fib[tot - 1];
		if (fib[tot + 1] > INF) break;
		tot++;
	}
	a[1].push_back(make_pair(1, 2));
	a[1].push_back(make_pair(1, 3));
	a[1].push_back(make_pair(1, 4));
	for (int i = 1; i <= tot - 3; i++) {
		for (unsigned j = 0; j < a[i].size(); j++) {
			long long x = a[i][j].first;
			long long y = a[i][j].second;
			x += y;
			while (x <= fib[i + 3] + fib[i]){
				a[i + 1].push_back(make_pair(y, x));
				x += y;
			}
		}
	}
	int q; read(q);
	while (q--){
		long long n, m;
		read(n), read(m);
		if (n > m) swap(n, m);
		int ans = 1;
		while (fib[ans + 1] <= n && fib[ans + 2] <= m)
			ans++;
		printf("%d ", ans);
		if (ans == 1) {
			printf("%lld\n", n * m % P);
			continue;
		}
		else {
			long long num = 0;
			for (unsigned j = 0; j < a[ans - 1].size(); j++) {
				long long x = a[ans - 1][j].first;
				long long y = a[ans - 1][j].second;
				if (x <= m && y <= n) num = (num + (m - x) / y) % P;
				if (x <= n && y <= m) num = (num + (n - x) / y) % P;
			}
			printf("%lld\n", num);
		}
	}
}
```