#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cstdlib>

#include <algorithm>
#include <numeric>
#include <limits>
#include <functional>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define LOG(FMT...) fprintf(stderr, INF)

#define fi first
#define se second
#define pb push_back
#define mp make_pair

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

const ll mod = 1e9 + 7;
const int F = 87, N = 105;

ll fib[N];
vector<P> t[N];

inline void upd(ll &a, ll b) {
  if ((a += b) >= mod)
    a -= mod;
}

int main() {
  t[1].pb(mp(1, 2));
  t[1].pb(mp(1, 3));
  t[1].pb(mp(1, 4));
  fib[0] = fib[1] = 1;
  for (int i = 2; i <= F; ++i)
    fib[i] = fib[i - 1] + fib[i - 2];
  for (int i = 1; i <= F; ++i)
    for (int j = 0; j < t[i].size(); ++j) {
      ll x = t[i][j].se, y = t[i][j].fi + x;
	  while (y <= fib[i] + fib[i + 3]) {
	    t[i + 1].pb(mp(x, y));
	    y += x;
      }
    }
  int q;
  scanf("%d", &q);
  while (q--) {
    ll n, m;
	scanf("%lld%lld", &n, &m);
	if (n > m)
	  swap(n, m);
	int p = 1;
	while (fib[p + 1] <= n && fib[p + 2] <= m)
	  ++p;
	printf("%d ", p);
	if (p == 1) {
	  printf("%lld\n", (ll)(n % mod) * (m % mod) % mod);
	  continue;
    }
    ll ans = 0;
    for (int j = 0; j < t[p - 1].size(); ++j) {
	  ll x = t[p - 1][j].fi, y = t[p - 1][j].se;
	  if (y <= n)
	    upd(ans, (m - x) / y % mod);
	  if (y <= m)
	    upd(ans, (n - x) / y % mod);
    }
    printf("%lld\n", ans);
  }
  return 0;
}


### Prompt

Generate 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 <cmath>
#include <ctime>
#include <cstring>
#include <cstdlib>

#include <algorithm>
#include <numeric>
#include <limits>
#include <functional>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <queue>
#define LOG(FMT...) fprintf(stderr, INF)

#define fi first
#define se second
#define pb push_back
#define mp make_pair

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

const ll mod = 1e9 + 7;
const int F = 87, N = 105;

ll fib[N];
vector<P> t[N];

inline void upd(ll &a, ll b) {
  if ((a += b) >= mod)
    a -= mod;
}

int main() {
  t[1].pb(mp(1, 2));
  t[1].pb(mp(1, 3));
  t[1].pb(mp(1, 4));
  fib[0] = fib[1] = 1;
  for (int i = 2; i <= F; ++i)
    fib[i] = fib[i - 1] + fib[i - 2];
  for (int i = 1; i <= F; ++i)
    for (int j = 0; j < t[i].size(); ++j) {
      ll x = t[i][j].se, y = t[i][j].fi + x;
	  while (y <= fib[i] + fib[i + 3]) {
	    t[i + 1].pb(mp(x, y));
	    y += x;
      }
    }
  int q;
  scanf("%d", &q);
  while (q--) {
    ll n, m;
	scanf("%lld%lld", &n, &m);
	if (n > m)
	  swap(n, m);
	int p = 1;
	while (fib[p + 1] <= n && fib[p + 2] <= m)
	  ++p;
	printf("%d ", p);
	if (p == 1) {
	  printf("%lld\n", (ll)(n % mod) * (m % mod) % mod);
	  continue;
    }
    ll ans = 0;
    for (int j = 0; j < t[p - 1].size(); ++j) {
	  ll x = t[p - 1][j].fi, y = t[p - 1][j].se;
	  if (y <= n)
	    upd(ans, (m - x) / y % mod);
	  if (y <= m)
	    upd(ans, (n - x) / y % mod);
    }
    printf("%lld\n", ans);
  }
  return 0;
}


```