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 <cstdio>
#include <cstring>
#include <algorithm>
#include <utility>
#define FR first
#define SE second
#define MOD 1000000007
#define inf (ll)(1e18)

using namespace std;

typedef long long ll;
typedef pair<ll,ll> pr;

ll f[310];
int mk;

pr a[310][310];
int num[310];

void build(int k) {
  for(int i=1;i<=num[k-1];i++) {
  	ll x=a[k-1][i].SE,y=a[k-1][i].FR;
  	y+=x;
  	for(;y<=f[k+2];y+=x) a[k][++num[k]]=pr(x,y);
  }
}

void pre() {
  f[0]=f[1]=1;
  for(mk=1;f[mk]<=inf;mk++) f[mk+1]=f[mk-1]+f[mk];
  mk--;
  num[1]=3;a[1][1]=pr(1,2);a[1][2]=pr(1,3);a[1][3]=pr(1,4);
  for(int i=2;i<=mk;i++) build(i);
}

inline ll solve(ll n,ll m,ll k) {
  return (m>n)?0:((n-m)/k+1)%MOD;
}

ll calc(ll n,ll m,int k) {
  ll ans=0;
  if (k==1) {
  	for(int i=1;i<=n;i++) ans=(ans+m)%MOD;
  }
  else {
  	for(int i=1;i<=num[k-1];i++) {
  		ll x=a[k-1][i].SE,y=a[k-1][i].FR;
  		if (x<=n) ans=(ans+solve(m,x+y,x))%MOD;
  		if (x<=m) ans=(ans+solve(n,x+y,x))%MOD;
	  }
  }
  return ans;
}

int main() {
  pre();
  int cases;
  scanf("%d",&cases);
  for(;cases;cases--) {
  	ll n=0,m=0;
  	scanf("%lld%lld",&n,&m);
  	if (n>m) swap(n,m);
  	int ans=1;
  	for(;n>=f[ans+1]&&m>=f[ans+2];ans++);
  	printf("%d %lld\n",ans,calc(n,m,ans));
  }
  return 0;
}