Problem: Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.

Help Bender to solve this difficult task.

Input

The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.

Output

If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.

If there are multiple solutions, print any of them.

Examples

Input

4 2
0 0
0 2
2 2
2 0
4 4


Output

YES
1 -1 2 -1 


Input

6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3


Output

YES
1 -1 2 -1 3 -1 


Input

6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3


Output

NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline bool isPrime(T n) {
  if (n <= 1) return false;
  for (T i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
template <class T>
inline T Mod(T n, T m) {
  return (n % m + m) % m;
}
template <class T>
string toString(T n) {
  ostringstream oss;
  oss << n;
  oss.flush();
  return oss.str();
}
int toInt(string s) {
  int r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
long long toLl(string s) {
  long long r = 0;
  istringstream sin(s);
  sin >> r;
  return r;
}
template <class T>
void debug(const T& e) {
  cout << e << endl;
}
template <class T1, class T2>
void debug(const T1& e1, const T2& e2) {
  cout << e1 << "\t" << e2 << endl;
}
template <class T1, class T2, class T3>
void debug(const T1& e1, const T2& e2, const T3& e3) {
  cout << e1 << "\t" << e2 << "\t" << e3 << endl;
}
template <class T1, class T2, class T3, class T4>
void debug(const T1& e1, const T2& e2, const T3& e3, const T4& e4) {
  cout << e1 << "\t" << e2 << "\t" << e3 << "\t" << e4 << endl;
}
template <class T1, class T2, class T3, class T4, class T5>
void debug(const T1& e1, const T2& e2, const T3& e3, const T4& e4,
           const T5& e5) {
  cout << e1 << "\t" << e2 << "\t" << e3 << "\t" << e4 << "\t" << e5 << endl;
}
template <class T>
void debug(vector<T>& e) {
  int i;
  for (i = 0; i < SZ(e); i++) cout << e[i] << " ";
  cout << endl;
}
template <class T>
void debug(vector<basic_string<T> >& e) {
  int i, j;
  for (i = 0; i < SZ(e); i++) {
    for (j = 0; j < SZ(e[i]); j++) cout << e[i][j];
    cout << endl;
  }
  cout << endl;
}
template <class T>
void debug(vector<vector<T> >& e, int row, int col) {
  int i, j;
  for (i = 0; i < row; i++) {
    for (j = 0; j < col; j++) cout << e[i][j] << "\t";
    cout << endl;
  }
  cout << endl;
}
template <class T>
void debug(T e[100][100], int row, int col) {
  int i, j;
  for (i = 0; i < row; i++) {
    for (j = 0; j < col; j++) cout << e[i][j] << " ";
    cout << endl;
  }
}
long long Pow(long long B, long long P) {
  long long R = 1;
  while (P > 0) {
    if (P % 2 == 1) R = (R * B);
    P /= 2;
    B = (B * B);
  }
  return R;
}
int l[10001], x[10001], y[10001], sett[10001], flag[10001];
int main() {
  int i, n, res, sum, j, m, dis, dis1, s, ss;
  while (scanf("%d%d", &n, &m) == 2) {
    s = ss = 0;
    scanf("%d%d", &x[0], &y[0]);
    sum = res = 0;
    for (i = 1; i < n; i++) {
      scanf("%d%d", &x[i], &y[i]);
      sum += (abs(x[i] - x[i - 1]) + abs(y[i] - y[i - 1]));
    }
    memset(sett, 0, sizeof(sett));
    memset(flag, 0, sizeof(flag));
    for (i = 0; i < m; i++) {
      scanf("%d", &l[i]);
      res += l[i];
    }
    if (res < sum)
      printf("NO\n");
    else {
      memset(sett, 0, sizeof(sett));
      memset(flag, 0, sizeof(flag));
      for (i = 0; i < n; i += 2) {
        if (i == 0) {
          dis = (abs(x[i] - x[i + 1]) + abs(y[i] - y[i + 1]));
          dis1 = (abs(x[i] - x[n - 1]) + abs(y[i] - y[n - 1]));
        } else {
          dis = (abs(x[i] - x[i + 1]) + abs(y[i] - y[i + 1]));
          dis1 = (abs(x[i] - x[i - 1]) + abs(y[i] - y[i - 1]));
        }
        dis = dis + dis1;
        for (j = 0; j < m; j++) {
          if (l[j] == dis && flag[j] == 0) {
            sett[i] = j + 1;
            flag[j] = 1;
            break;
          }
        }
        if (j == m) {
          s = 1;
          break;
        }
      }
      if (s == 0) {
        printf("YES\n");
        if (sett[0] > 0)
          printf("%d", sett[0]);
        else
          printf("-1");
        for (i = 1; i < n; i++) {
          if (sett[i] > 0)
            printf(" %d", sett[i]);
          else
            printf(" -1");
        }
        printf("\n");
      } else {
        memset(sett, 0, sizeof(sett));
        memset(flag, 0, sizeof(flag));
        for (i = 1; i < n; i += 2) {
          if (i == n - 1) {
            dis = (abs(x[i] - x[i - 1]) + abs(y[i] - y[i - 1]));
            dis1 = (abs(x[i] - x[0]) + abs(y[i] - y[0]));
          } else {
            dis = (abs(x[i] - x[i + 1]) + abs(y[i] - y[i + 1]));
            dis1 = (abs(x[i] - x[i - 1]) + abs(y[i] - y[i - 1]));
          }
          dis = dis + dis1;
          for (j = 0; j < m; j++) {
            if (l[j] == dis && flag[j] == 0) {
              sett[i] = j + 1;
              flag[j] = 1;
              break;
            }
          }
          if (j == m) {
            ss = 1;
            break;
          }
        }
        if (ss == 0) {
          printf("YES\n");
          if (sett[0] > 0)
            printf("%d", sett[0]);
          else
            printf("-1");
          for (i = 1; i < n; i++) {
            if (sett[i] > 0)
              printf(" %d", sett[i]);
            else
              printf(" -1");
          }
          printf("\n");
        }
      }
      if (s == 1 && ss == 1) printf("NO\n");
    }
  }
  return 0;
}