Educational Codeforces Round 52 (Rated for Div. 2)

There is a special offer in Vasya's favourite supermarket: if the customer buys $aa chocolate bars, he or she may take bb additional bars for free. This special offer can be used any number of times.$

Vasya currently has $ss roubles, and he wants to get as many chocolate bars for free. Each chocolate bar costs cc roubles. Help Vasya to calculate the maximum possible number of chocolate bars he can get!$

Input

The first line contains one integer $tt (1\leqt\leq1001\leqt\leq100) — the number of testcases.$

Each of the next $tt lines contains four integers s,a,b,c (1\leqs,a,b,c\leq109)s,a,b,c (1\leqs,a,b,c\leq109) — the number of roubles Vasya has, the number of chocolate bars you have to buy to use the special offer, the number of bars you get for free, and the cost of one bar, respectively.$

Output

Print $tt lines. ii-th line should contain the maximum possible number of chocolate bars Vasya can get in ii-th test.$

Example

input

Copy

2
10 3 1 1
1000000000 1 1000000000 1

output

Copy

13
1000000001000000000

Note

In the first test of the example Vasya can buy $99 bars, get 33 for free, buy another bar, and so he will get 1313 bars.$

In the second test Vasya buys $10000000001000000000 bars and gets 10000000000000000001000000000000000000 for free. So he has 10000000010000000001000000001000000000 bars.$

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 
 6 int main()
 7 {
 8     long long t,s,a,b,c;
 9     scanf("%lld",&t);
10     while(t--){
11         scanf("%lld %lld %lld %lld",&s,&a,&b,&c);
12         long long temp=a*c;
13         if(s<temp){
14             printf("%lld\n",s/c);
15         }else{
16             long long cou=(s/temp)*(a+b)+(s%temp)/c;
17             printf("%lld\n",cou);
18         }
19     }
20     return 0;
21 }

Vasya has got an undirected graph consisting of $nn vertices and mm edges. This graph doesn't contain any self-loops or multiple edges. Self-loop is an edge connecting a vertex to itself. Multiple edges are a pair of edges such that they connect the same pair of vertices. Since the graph is undirected, the pair of edges (1,2)(1,2) and (2,1)(2,1) is considered to be multiple edges. Isolated vertex of the graph is a vertex such that there is no edge connecting this vertex to any other vertex.$

Vasya wants to know the minimum and maximum possible number of isolated vertices in an undirected graph consisting of $nn vertices and mm edges.$

Input

The only line contains two integers $nn and m (1\leqn\leq105,0\leqm\leqn(n-1)2)m (1\leqn\leq105,0\leqm\leqn(n-1)2).$

It is guaranteed that there exists a graph without any self-loops or multiple edges with such number of vertices and edges.

Output

In the only line print two numbers $minmin and maxmax — the minimum and maximum number of isolated vertices, respectively.$

Examples

input

Copy

4 2

output

Copy

0 1

input

Copy

3 1

output

Copy

1 1

Note

In the first example it is possible to construct a graph with $00 isolated vertices: for example, it should contain edges (1,2)(1,2) and (3,4)(3,4). To get one isolated vertex, we may construct a graph with edges (1,2)(1,2) and (1,3)(1,3).$

In the second example the graph will always contain exactly one isolated vertex.

wrong到结束......

还是不对

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 
 6 typedef long long ll;
 7 
 8 ll cc[100005];
 9 
10 int main()
11 {
12     ll n,m;
13     for(ll i=1;i<=100004;i++){
14         cc[i]=i*(i-1)/2;
15     }
16     while(~scanf("%lld %lld",&n,&m)){
17         ll minn,maxx;
18         if(n==1){
19             printf("1 1\n");
20             continue;
21         }
22         if(n==0){
23             printf("0 0\n");
24             continue;
25         }
26         if(m==0){
27             printf("%lld %lld\n",n,n);
28             continue;
29         }
30         if(m*2>=n){
31             minn=0;
32             if(m>cc[n-1]){
33                 maxx=0;
34             }else{
35                 for(int i=1;i<=99999;i++){
36                     if(m>cc[i]&&m<=cc[i+1]){
37                         maxx=n-(i+1);
38                         break;
39                     }
40                 }
41 
42             }
43         }else{
44             minn=n-2*m;
45             maxx=n-m-1;
46         }
47         printf("%lld %lld\n",minn,maxx);
48     }
49     return 0;
50 }