Problem: Monocarp would like to open a bakery in his local area. But, at first, he should figure out whether he can compete with other shops.

Monocarp plans that the bakery will work for n days. On the i-th day, a_i loaves of bread will be baked in the morning before the opening. At the end of the n-th day, Monocarp will sell all the remaining bread that wasn't sold earlier with a huge discount.

Because of how bread is stored, the bakery seller sells the bread in the following order: firstly, he sells the loaves that were baked that morning; secondly, he sells the loaves that were baked the day before and weren't sold yet; then the loaves that were baked two days before and weren't sold yet, and so on. That's why some customers may buy a rather stale bread and will definitely spread negative rumors.

Let's define loaf spoilage as the difference between the day it was baked and the day it was sold. Then the unattractiveness of the bakery will be equal to the maximum spoilage among all loaves of bread baked at the bakery.

Suppose Monocarp's local area has consumer demand equal to k, it means that each day k customers will come to the bakery and each of them will ask for one loaf of bread (the loaves are sold according to the aforementioned order). If there is no bread left, then the person just doesn't buy anything. During the last day sale, all the remaining loaves will be sold (and they will still count in the calculation of the unattractiveness).

Monocarp analyzed his competitors' data and came up with m possible consumer demand values k_1, k_2, ..., k_m, and now he'd like to calculate the unattractiveness of the bakery for each value of demand. Can you help him?

Input

The first line contains two integers n and m (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ m ≤ 2 ⋅ 10^5) — the number of days the bakery is open and the number of possible values of consumer demand.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of bread loaves that will be baked each day.

The third line contains m integers k_1, k_2, ..., k_m (1 ≤ k_1 < k_2 < ... < k_m ≤ 10^9) — the possible consumer demand values in the ascending order.

Output

Print m integers: for each consumer demand, print the unattractiveness of the bakery.

Examples

Input


5 4
5 2 1 3 7
1 3 4 10


Output


4 2 1 0 


Input


8 9
3 1 4 1 5 9 2 6
1 2 3 4 5 6 7 8 9


Output


7 5 3 3 2 1 1 1 0 

Note

In the first example, let's describe what happens for couple consumer demands:

If consumer demand is equal to 1: 

  * at day 1: 5 loaves are baked and only 1 is sold with spoilage equal to 1 - 1 = 0; 
  * at day 2: 4 loaves are left and 2 more are baked. Only 1 loaf was sold and it was the loaf baked today with spoilage 2 - 2 = 0; 
  * at day 3: 4 loaves from the first day and 1 loaf from the second day left. One more loaf was baked and was sold this day with spoilage 3 - 3 = 0; 
  * at day 4: 4 loaves from the first day and 1 loaf from the second day left. 3 more loaves were baked and one of them was sold this day with spoilage 4 - 4 = 0; 
  * at day 5: 4 loaves from the first day, 1 loaf from the second day and 2 loaves from the fourth day left. 7 more loaves were baked and, since it's the last day, all 14 loaves were sold. 4 loaves from the first day have the maximum spoilage equal to 5 - 1 = 4. 

In total, the unattractiveness of the bakery will be equal to 4.

If consumer demand is equal to 10 then all baked bread will be sold in the day it was baked and will have spoilage equal to 0.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define rint register int
#define rll register ll 
#define pii pair<int, int>
#define pll pair<ll, ll>
//#define p1 first
//#define p2 second
#define fors(i, a, b) for (register int i = (a); i <= (b); ++i)
#define _fors(i, a, b) for (ll i = (a); i >= (b); --i)
#define mp(a, b) make_pair(a, b)
#define mt(a, b, c) make_tuple(a, b, c)
#define mem(A, b) memset(A, b, sizeof(A))
#define all(X) (X).begin(), (X).end()
#define pb push_back

static struct FastInput {
  static constexpr int BUF_SIZE = 1 << 20;
  char buf[BUF_SIZE];
  size_t chars_read = 0;
  size_t buf_pos = 0;
  FILE *in = stdin;
  char cur = 0;
 
  inline char get_char() {
    if (buf_pos >= chars_read) {
      chars_read = fread(buf, 1, BUF_SIZE, in);
      buf_pos = 0;
      buf[0] = (chars_read == 0 ? -1 : buf[0]);
    }
    return cur = buf[buf_pos++];
  }
 
  inline void tie(int) {}
 
  inline explicit operator bool() {
    return cur != -1;
  }
 
  inline static bool is_blank(char c) {
    return c <= ' ';
  }
 
  inline bool skip_blanks() {
    while (is_blank(cur) && cur != -1) {
      get_char();
    }
    return cur != -1;
  }
 
  inline FastInput& operator>>(char& c) {
    skip_blanks();
    c = cur;
    return *this;
  }
  
  inline FastInput& operator>>(string& s) {
    if (skip_blanks()) {
      s.clear();
      do {
        s += cur;
      } while (!is_blank(get_char()));
    }
    return *this;
  }
 
  template <typename T>
  inline FastInput& read_integer(T& n) {
    // unsafe, doesn't check that characters are actually digits
    n = 0;
    if (skip_blanks()) {
      int sign = +1;
      if (cur == '-') {
        sign = -1;
        get_char();
      }
      do {
        n += n + (n << 3) + cur - '0';
      } while (!is_blank(get_char()));
      n *= sign;
    }
    return *this;
  }
 
  template <typename T>
  inline typename enable_if<is_integral<T>::value, FastInput&>::type operator>>(T& n) {
    return read_integer(n);
  }
  
  #if !defined(_WIN32) || defined(_WIN64)
  inline FastInput& operator>>(__int128& n) {
    return read_integer(n);
  }
  #endif
 
  template <typename T>
  inline typename enable_if<is_floating_point<T>::value, FastInput&>::type operator>>(T& n) {
    // not sure if really fast, for compatibility only
    n = 0;
    if (skip_blanks()) {
      string s;
      (*this) >> s;
      sscanf(s.c_str(), "%lf", &n);
    }
    return *this;
  }
} fast_input;
 
#define cin fast_input
 
static struct FastOutput {
  static constexpr int BUF_SIZE = 1 << 20;
  char buf[BUF_SIZE];
  size_t buf_pos = 0;
  static constexpr int TMP_SIZE = 1 << 20;
  char tmp[TMP_SIZE];
  FILE *out = stdout;
 
  inline void put_char(char c) {
    buf[buf_pos++] = c;
    if (buf_pos == BUF_SIZE) {
      fwrite(buf, 1, buf_pos, out);
      buf_pos = 0;
    }
  }
 
  ~FastOutput() {
    fwrite(buf, 1, buf_pos, out);
  }
 
  inline FastOutput& operator<<(char c) {
    put_char(c);
    return *this;
  }
 
  inline FastOutput& operator<<(const char* s) {
    while (*s) {
      put_char(*s++);
    }
    return *this;
  }
 
  inline FastOutput& operator<<(const string& s) {
    for (int i = 0; i < (int) s.size(); i++) {
      put_char(s[i]);
    }
    return *this;
  }
 
  template <typename T>
  inline char* integer_to_string(T n) {
    // beware of TMP_SIZE
    char* p = tmp + TMP_SIZE - 1;
    if (n == 0) {
      *--p = '0';
    } else {
      bool is_negative = false;
      if (n < 0) {
        is_negative = true;
        n = -n;
      }
      while (n > 0) {
        *--p = (char) ('0' + n % 10);
        n /= 10;
      }
      if (is_negative) {
        *--p = '-';
      }
    }
    return p;
  }
 
  template <typename T>
  inline typename enable_if<is_integral<T>::value, char*>::type stringify(T n) {
    return integer_to_string(n);
  }
 
  #if !defined(_WIN32) || defined(_WIN64)
  inline char* stringify(__int128 n) {
    return integer_to_string(n);
  }
  #endif
 
  template <typename T>
  inline typename enable_if<is_floating_point<T>::value, char*>::type stringify(T n) {
    sprintf(tmp, "%.17f", n);
    return tmp;
  }
 
  template <typename T>
  inline FastOutput& operator<<(const T& n) {
    auto p = stringify(n);
    for (; *p != 0; p++) {
      put_char(*p);
    }
    return *this;
  }
} fast_output;

#define cout fast_output
#define endl '\n'
#define int long long

const int _ = 2e5 + 15;
set<pll> s1;
set<int> s2;
multiset<int> s3;
int a[_], s[_], q[_], v[_], ans[_];
int n, m;

signed main() {
  //freopen("in.txt", "r", stdin);
  //freopen("out.txt", "w", stdout);
  ios::sync_with_stdio(false);
  cin.tie(0);
  int start_time = clock();
  /*
  int T;
  cin >> T;
  while (T--) {
    
  }*/
  cin >> n >> m;
  fors(i, 1, n) cin >> a[i];
  a[n] = -1e18;
  fors(i, 1, n) s[i] = s[i-1] + a[i];
  fors(i, 1, m) cin >> q[i];
  fors(i, 1, n) s1.insert(mp(-a[i] + 1, i));
  fors(i, 0, n) s2.insert(i);
  fors(i, 1, n) s3.insert(1);
  fors(i, 1, n) v[i] = a[i] - 1;
  _fors(i, m, 1) {
    while (!s1.empty() && -s1.begin()->first >= q[i]) {
      pll now = *s1.begin();
      auto it = s2.find(now.second), it2 = it, it3 = it;
      it2--, it3++;
      int x = *it2, y = *it3, z = *it;
      s3.erase(s3.find(z - x));
      s3.erase(s3.find(y - z));
      s3.insert(y - x);
      s1.erase(s1.begin());
      s1.erase(mp(-v[y], y));
      v[y] = (s[y] - s[x] - 1) / (y - x);
      s1.insert(mp(-v[y], y));
      s2.erase(z);
    }
    auto it = s3.end();
    it--;
    ans[i] = *it;
  }
  fors(i, 1, m) cout << ans[i] - 1 << ' ';
  cout << endl;
  //cout << "time = " << clock() - start_time << endl;
  return 0;
}