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 <iostream>

#include <cstdio>

#include <vector>

#include <cstring>

#include <cstdlib>

#include <cassert>

#include <iostream>

#define rep(i,a,b) for(int i = a; i <= b; i++)

#define dep(i,a,b) for(int i = a; i >= b; i--)

#define Rep(i,a) for(int i = 0; i < a; i++)

#define pb push_back

#define mp make_pair

typedef long long LL;

using namespace std;

const int N = 90, mod = 1e9 + 7;
typedef pair<LL, LL> pii;
LL f[N];
int cnt = 0;
vector<pii> g[N];
void dfs(int t, LL a, LL b) {
	if (a > 2LL * f[t] || b > LL(1e18)) return;
	if (f[t] <= a && a < f[t + 2] && f[t + 1] <= b && b < f[t + 2]) g[t].pb(mp(a, b));
	rep(i,1,4) dfs(t + 1, b, a + b * i);
}

LL read() {
	char c = getchar(); LL x = 0;
	while (c < '0' || c > '9') c = getchar();
	while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x;
}


int main() {

//	freopen("gcd.in","r",stdin);

//	freopen("gcd.out","w",stdout);

	f[0] = f[1] = 1; rep(i,2,87) f[i] = f[i - 1] + f[i - 2];
	dfs(0, 1, 1);
	int Q; scanf("%d",&Q);
	while (Q--) {
		LL a, b; a = read(), b = read();
		if (a > b) swap(a, b); int mx = 1;
		dep(i,86,1) if (f[i] <= a && f[i + 1] <= b) { mx = i; break; }
		if (mx == 1) printf("%d %d\n",mx, 1LL * a * b % mod);
		else {
			int len = g[mx].size(); int ans = 0;
			Rep(j,len) {
				LL x = g[mx][j].first, y = g[mx][j].second;
				if (x <= a && y <= b) ans++;
				if (y <= a && x <= b) ans++;
			}
			len = g[mx - 1].size();
			Rep(j,len) {
				LL x = g[mx - 1][j].first, y = g[mx - 1][j].second;
				if (y <= a && b >= f[mx + 2]) 
					ans = (ans + (b - x) / y - (f[mx + 2] - 1 - x) / y) % mod;
			}
			if (ans < 0) ans += mod;
			printf("%d %d\n",mx, ans);
		}
	}
	return 0;

}