Problem: Sereja has an array a, consisting of n integers a1, a2, ..., an. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out m integers l1, l2, ..., lm (1 ≤ li ≤ n). For each number li he wants to know how many distinct numbers are staying on the positions li, li + 1, ..., n. Formally, he want to find the number of distinct numbers among ali, ali + 1, ..., an.?

Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each li.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the array elements.

Next m lines contain integers l1, l2, ..., lm. The i-th line contains integer li (1 ≤ li ≤ n).

Output

Print m lines — on the i-th line print the answer to the number li.

Examples

Input

10 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10


Output

6
6
6
6
6
5
4
3
2
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using namespace std;
int nxt() {
  int x;
  cin >> x;
  return x;
}
const long long INF = int(1e9) + 99;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
const int N = int(2e5) + 99;
vector<int> plist;
void SieveOfEratosthenes(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) plist.push_back(p);
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int _mergeSort(int arr[], int temp[], int left, int right);
int merge(int arr[], int temp[], int left, int mid, int right);
int mergeSort(int arr[], int array_size) {
  int temp[array_size];
  return _mergeSort(arr, temp, 0, array_size - 1);
}
int _mergeSort(int arr[], int temp[], int left, int right) {
  int mid, inv_count = 0;
  if (right > left) {
    mid = (right + left) / 2;
    inv_count += _mergeSort(arr, temp, left, mid);
    inv_count += _mergeSort(arr, temp, mid + 1, right);
    inv_count += merge(arr, temp, left, mid + 1, right);
  }
  return inv_count;
}
int merge(int arr[], int temp[], int left, int mid, int right) {
  int i, j, k;
  int inv_count = 0;
  i = left;
  j = mid;
  k = left;
  while ((i <= mid - 1) && (j <= right)) {
    if (arr[i] <= arr[j]) {
      temp[k++] = arr[i++];
    } else {
      temp[k++] = arr[j++];
      inv_count = inv_count + (mid - i);
    }
  }
  while (i <= mid - 1) temp[k++] = arr[i++];
  while (j <= right) temp[k++] = arr[j++];
  for (i = left; i <= right; i++) arr[i] = temp[i];
  return inv_count;
}
long long binomialCoeff(long long n, long long k) {
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
int MAXN = 8e5;
pair<int, int> t[800000];
pair<int, int> combine(pair<int, int> a, pair<int, int> b) {
  if (a.first < b.first) return a;
  if (b.first < a.first) return b;
  return make_pair(a.first, ((a.second > b.second) ? b.second : a.second));
}
void build(int a[], int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = make_pair(a[tl], tl);
  } else {
    int tm = (tl + tr) / 2;
    build(a, v * 2, tl, tm);
    build(a, v * 2 + 1, tm + 1, tr);
    t[v] = combine(t[v * 2], t[v * 2 + 1]);
  }
}
pair<int, int> get_first(int v, int lv, int rv, int l, int r, int x) {
  if (lv > r || rv < l) return make_pair(INF, -1);
  if (l <= lv && rv <= r) {
    if (t[v].first >= x) return make_pair(INF, -1);
    while (lv != rv) {
      int mid = lv + (rv - lv) / 2;
      if (t[2 * v].first < x) {
        v = 2 * v;
        rv = mid;
      } else {
        v = 2 * v + 1;
        lv = mid + 1;
      }
    }
    return make_pair(x - 1, lv);
  }
  int mid = lv + (rv - lv) / 2;
  pair<int, int> rs = get_first(2 * v, lv, mid, l, r, x);
  if (rs.second != -1) return rs;
  return get_first(2 * v + 1, mid + 1, rv, l, r, x);
}
void solve() {
  long long l, r, g, x, y, m, n, ans1 = 1e9 + 7, ans13 = 0, ans2 = 0, ans3 = 0,
                                 value, ans4 = 0, flag1 = 0, ans5 = 0,
                                 ans6 = -1e18, ans = 0, k = 0, d;
  cin >> n >> m;
  set<int> s;
  vector<int> v(n), v1(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  ;
  for (int i = n - 1; i >= 0; i--) {
    s.insert(v[i]);
    v1[i] = s.size();
  }
  for (long long i = 0; i < m; i++) {
    cin >> x;
    cout << v1[x - 1] << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  SieveOfEratosthenes(100000);
  long long t = 1;
  while (t--) {
    solve();
    cout << "\n";
    cerr << t << " ";
  }
  return 0;
}