You are given an array a consisting of n integer numbers.

You have to color this array in k colors in such a way that: 

  * Each element of the array should be colored in some color; 
  * For each i from 1 to k there should be at least one element colored in the i-th color in the array; 
  * For each i from 1 to k all elements colored in the i-th color should be distinct. 



Obviously, such coloring might be impossible. In this case, print "NO". Otherwise print "YES" and any coloring (i.e. numbers c_1, c_2, ... c_n, where 1 ≤ c_i ≤ k and c_i is the color of the i-th element of the given array) satisfying the conditions above. If there are multiple answers, you can print any.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 5000) — the length of the array a and the number of colors, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 5000) — elements of the array a.

Output

If there is no answer, print "NO". Otherwise print "YES" and any coloring (i.e. numbers c_1, c_2, ... c_n, where 1 ≤ c_i ≤ k and c_i is the color of the i-th element of the given array) satisfying the conditions described in the problem statement. If there are multiple answers, you can print any.

Examples

Input


4 2
1 2 2 3


Output


YES
1 1 2 2


Input


5 2
3 2 1 2 3


Output


YES
2 1 1 2 1


Input


5 2
2 1 1 2 1


Output


NO

Note

In the first example the answer 2~ 1~ 2~ 1 is also acceptable.

In the second example the answer 1~ 1~ 1~ 2~ 2 is also acceptable.

There exist other acceptable answers for both examples.
#include <bits/stdc++.h>
using namespace std;
void sor(vector<long long> &x) { sort(x.begin(), x.end()); }
void rev(vector<long long> &x) { reverse(x.begin(), x.end()); }
long long gcd(long long x, long long y) {
  if (y == 0) {
    return x;
  }
  return gcd(y, x % y);
}
long long max(long long x, long long y) {
  if (x > y) {
    return x;
  } else {
    return y;
  }
}
long long min(long long x, long long y) {
  if (x > y) {
    return y;
  } else {
    return x;
  }
}
void primeFactors(long long n, vector<long long> &vec) {
  while (n % 2 == 0) {
    vec.push_back(2);
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      vec.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) vec.push_back(n);
}
long long lcm(long long x, long long y) { return x * y / gcd(x, y); }
long long g(long long x) {
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) {
      return i;
    }
  }
  return 0;
}
long long bina(vector<long long> a, long long n, long long m) {
  long long start = 0, end = n - 1;
  long long c = 0;
  if (m >= a[n - 1]) {
    return n;
  }
  while (start <= end) {
    long long mid = (start + end) / 2;
    if (m >= a[mid]) {
      c = mid + 1;
      start = mid + 1;
    } else {
      end = mid - 1;
    }
  }
  return c;
}
long long binarySearch(vector<long long> arr, long long l, long long r,
                       long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
int subArraylen(long long arr[], long long n, long long K) {
  unordered_map<int, int> mp;
  mp[arr[0]] = 0;
  for (int i = 1; i < n; i++) {
    arr[i] = arr[i] + arr[i - 1];
    mp[arr[i]] = i;
  }
  int len = INT_MAX;
  for (int i = 0; i < n; i++) {
    if (arr[i] < K)
      continue;
    else {
      int x = arr[i] - K;
      if (x == 0) len = min(len, i);
      if (mp.find(x) == mp.end())
        continue;
      else {
        len = min(len, i - mp[x]);
      }
    }
  }
  return len;
}
long long prime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return 0;
}
long long nextPowerOf2(long long n) {
  long long p = 1;
  if (n && !(n & (n - 1))) return n;
  while (p < n) p <<= 1;
  return p;
}
long long countBits(long long number) { return (long long)log2(number) + 1; }
long long comb(long long c, long long n) {
  long long num = 1, d = 1;
  for (long long i = c; i > max(n, c - n); i--) {
    num *= i;
  }
  for (long long i = 2; i <= min(n, c - n); i++) {
    d *= i;
  }
  return (num / d);
}
long long countDivisors(long long n) {
  long long cnt = 0;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
int getCount(string a, string b) {
  if (b.length() % a.length() != 0) return -1;
  int count = b.length() / a.length();
  string str = "";
  for (int i = 0; i < count; i++) {
    str = str + a;
  }
  if (str == b) return count;
  return -1;
}
long long fact(long long n) {
  long long res = 1;
  for (int i = 2; i <= n; i++) res = res * i;
  return res;
}
long long nCr(long long n, long long r) {
  return fact(n) / (fact(r) * fact(n - r));
}
long long no(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long x = no(a, b / 2);
  if (b % 2) {
    return x * x * a;
  } else {
    return x * x;
  }
}
int countFreq(string &pat, string &txt) {
  int M = pat.length();
  int N = txt.length();
  int res = 0;
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (txt[i + j] != pat[j]) break;
    if (j == M) {
      res++;
      j = 0;
    }
  }
  return res;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  vector<long long> b(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sor(b);
  vector<long long> cc;
  long long x = 1;
  for (int i = 1; i < n; i++) {
    if (b[i] == b[i - 1]) {
      x++;
    } else {
      cc.push_back(x);
      x = 1;
    }
  }
  cc.push_back(x);
  sor(cc);
  if (cc[cc.size() - 1] > k) {
    cout << "NO";
    return;
  }
  long long j = 0;
  vector<long long> vec;
  long long o = 1;
  set<long long> s;
  long long start = 0;
  long long c = 0;
  long long i = 0;
  for (; i < n; i++) {
    long long l1 = s.size();
    s.insert(a[i]);
    long long l2 = s.size();
    if (l1 == l2) {
      for (int j = start; j < i; j++) {
        vec.push_back(o);
        o++;
        c++;
      }
      start = i;
      s.clear();
      s.insert(a[i]);
    }
  }
  if (c != n) {
    for (int j = start; j < i; j++) {
      vec.push_back(o);
      o++;
      c++;
    }
  }
  vector<long long> ans(n, 0);
  cout << "YES\n";
  long long mark = 1;
  for (long long i = 0; i < n; ++i) {
    long long p = a[i];
    if (a[i] != -1) {
      for (long long j = 0; j < n; ++j) {
        if (p == a[j]) {
          ans[j] = (mark % k) + 1;
          mark++;
          a[j] = -1;
        }
      }
    }
  }
  for (long long i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }
}
vector<long long> a[1000001];
long long vis[1000001];
long long col[1000001];
bool dfs(long long x, long long c) {
  vis[x] = 1;
  col[x] = c;
  for (long long child : a[x]) {
    if (vis[child] == 0) {
      bool re = dfs(child, c ^ 1);
      if (re == false) {
        return false;
      }
    } else {
      if (col[x] == col[child]) {
        return false;
      }
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long t = 1;
  for (long long tt = 0; tt < t; ++tt) {
    solve();
  }
  return 0;
}
