Vasya goes to visit his classmate Petya. Vasya knows that Petya’s apartment number is n.
There is only one entrance in Petya’s house and the distribution of apartments is the following: the first floor contains 2 apartments, every other floor contains x apartments each. Apartments are numbered starting from one, from the first floor. I.e. apartments on the first floor have numbers 1 and 2, apartments on the second floor have numbers from 3 to (x+2), apartments on the third floor have numbers from (x+3) to (2⋅x+2), and so on.
Your task is to find the number of floor on which Petya lives. Assume that the house is always high enough to fit at least n apartments.
You have to answer t independent test cases.
Input The first line of the input contains one integer t (1≤t≤1000) — the number of test cases. Then t test cases follow.
The only line of the test case contains two integers n and x (1≤n,x≤1000) — the number of Petya’s apartment and the number of apartments on each floor of the house except the first one (there are two apartments on the first floor).
Output For each test case, print the answer: the number of floor on which Petya lives.
Example Input 4 7 3 1 5 22 5 987 13 Output 3 1 5 77
这是一道思维题，已知房间号 n,和除第一层之外每层房间数x，求 Petya 住的楼层数，由题可知，第一层永远只有两间房，在第一层特殊处理 n-2,其余的层数为 n-x,直到n<=0为止。楼层范围小于1000
#include<cstdio> #include<iostream> using namespace std; int main() { int i,n,x,t; cin>>t; while(t--) { cin>>n>>x; for(i=1;i<=1005;i++) { if(i==1) { n-=2; } else { n-=x; } if(n<=0) { cout<<i<<endl; break; } } } return 0; }