B. A Leapfrog in the Array

http://codeforces.com/problemset/problem/949/B

Dima is a beginner programmer. During his working process, he regularly has to repeat the following operation again and again: to remove every second element from the array. One day he has been bored with easy solutions of this problem, and he has come up with the following extravagant algorithm.

Let's consider that initially array contains n numbers from 1 to n and the number i is located in the cell with the index 2i - 1 (Indices are numbered starting from one) and other cells of the array are empty. Each step Dima selects a non-empty array cell with the maximum index and moves the number written in it to the nearest empty cell to the left of the selected one. The process continues until all n numbers will appear in the first n cells of the array. For example if n = 4, the array is changing as follows:

You have to write a program that allows you to determine what number will be in the cell with index x(1 ≤ x ≤ n) after Dima's algorithm finishes.

Input

The first line contains two integers n and q (1 ≤ n ≤ 1018, 1 ≤ q ≤ 200 000), the number of elements in the array and the number of queries for which it is needed to find the answer.

Next q lines contain integers xi (1 ≤ xi ≤ n), the indices of cells for which it is necessary to output their content after Dima's algorithm finishes.

Output

For each of q queries output one integer number, the value that will appear in the corresponding array cell after Dima's algorithm finishes.

Examples

input

Copy

4 3
2
3
4

output

3
2
4

input

Copy

13 4
10
5
4
8

output

13
3
8
9

Note

The first example is shown in the picture.

In the second example the final array is [1, 12, 2, 8, 3, 11, 4, 9, 5, 13, 6, 10, 7].

 思维题，找规律

1.n为奇数的话，可以用n-1偶数过渡过来（只需要把最后一个丢到最前面）

2.询问的k为奇数的话，直接返回(k+1)/2，因为位置是不改变的

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 2147483647
const ll INF = 0x3f3f3f3f3f3f3f3fll;
#define ri register int
template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
template <class T> inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T> inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
#define scanf1(x) scanf("%d", &x)
#define scanf2(x, y) scanf("%d%d", &x, &y)
#define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X)
#define pi acos(-1)
#define me(x, y) memset(x, y, sizeof(x));
#define For(i, a, b) for (int i = a; i <= b; i++)
#define FFor(i, a, b) for (int i = a; i >= b; i--)
#define bug printf("***********
");
#define mp make_pair
#define pb push_back
const int maxn = 10005;
// name*******************************
ll T;
ll k, n;
ll ans;
// function******************************
ll solve(ll n, ll k) {
  if (n % 2 == 0) {
    if (k % 2) {
      return (k + 1) / 2;
    } else {
      return solve(n / 2, k / 2) + n / 2;
    }
  } else {
    if (k % 2) {
      return (k + 1) / 2;
    } else {
      if (k == 2) {
        return solve(n - 1, n - 1) + 1;
      } else {
        return solve(n - 1, k - 2) + 1;
      }
    }
  }
}

//***************************************
int main() {
  //    ios::sync_with_stdio(0);
  //    cin.tie(0);
  // freopen("test.txt", "r", stdin);
  //  freopen("outout.txt","w",stdout);
  cin >> n >> T;
  while (T--) {
    ans = 0;
    ll k;
    cin >> k;
    cout <<solve(n, k) << endl;
  }

  return 0;
}