Problem: Polycarp is an experienced participant in Codehorses programming contests. Now he wants to become a problemsetter.

He sent to the coordinator a set of n problems. Each problem has it's quality, the quality of the i-th problem is ai (ai can be positive, negative or equal to zero). The problems are ordered by expected difficulty, but the difficulty is not related to the quality in any way. The easiest problem has index 1, the hardest problem has index n.

The coordinator's mood is equal to q now. After reading a problem, the mood changes by it's quality. It means that after the coordinator reads a problem with quality b, the value b is added to his mood. The coordinator always reads problems one by one from the easiest to the hardest, it's impossible to change the order of the problems.

If after reading some problem the coordinator's mood becomes negative, he immediately stops reading and rejects the problemset.

Polycarp wants to remove the minimum number of problems from his problemset to make the coordinator's mood non-negative at any moment of time. Polycarp is not sure about the current coordinator's mood, but he has m guesses "the current coordinator's mood q = bi".

For each of m guesses, find the minimum number of problems Polycarp needs to remove so that the coordinator's mood will always be greater or equal to 0 while he reads problems from the easiest of the remaining problems to the hardest.

Input

The first line of input contains two integers n and m (1 ≤ n ≤ 750, 1 ≤ m ≤ 200 000) — the number of problems in the problemset and the number of guesses about the current coordinator's mood.

The second line of input contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the qualities of the problems in order of increasing difficulty.

The third line of input contains m integers b1, b2, ..., bm (0 ≤ bi ≤ 1015) — the guesses of the current coordinator's mood q.

Output

Print m lines, in i-th line print single integer — the answer to the problem with q = bi.

Example

Input

6 3
8 -5 -4 1 -7 4
0 7 3


Output

2
0
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
template <class T>
void dbv(T a) {
  for (auto x : a) {
    cout << x << " ** ";
  }
  cout << endl;
}
long long calc(long long in, vector<long long> &a) {
  priority_queue<long long> pq;
  long long cum = in;
  long long cnt = 0;
  long long n = a.size();
  for (int i = (0); i < (n); ++i) {
    pq.push(-a[i]);
    cum += a[i];
    if (cum < 0) {
      cum += pq.top();
      pq.pop();
      cnt++;
    }
  }
  return cnt;
}
void output() {
  long long n, q;
  cin >> n >> q;
  vector<long long> a(n);
  for (int i = (0); i < (n); ++i) {
    cin >> a[i];
  }
  long long st = calc(0, a);
  vector<long long> lims;
  lims.push_back(0);
  for (int val = (st - 1); val >= (0); --val) {
    long long low = lims[st - val - 1] + 1;
    long long high = LINF;
    while (low < high) {
      long long mid = (low + high) / 2;
      if (calc(mid, a) <= val)
        high = mid;
      else
        low = mid + 1;
    }
    lims.push_back(low);
  }
  lims.push_back(1e18);
  for (int i = (0); i < (q); ++i) {
    long long qu;
    cin >> qu;
    long long low = 0;
    long long high = st + 1;
    while (low < high) {
      long long mid = (low + high) / 2;
      if (qu < lims[mid])
        high = mid;
      else
        low = mid + 1;
    }
    cout << (st - (low - 1)) << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  for (int i = (0); i < (t); ++i) {
    output();
  }
  return 0;
}