Problem: There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.

Input

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.

Output

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.

Examples

Input

4 4
1 2 3 4


Output

YES
1
1 4
1 2 4
1 2 3 4


Input

5 2
3 2 4 1 3


Output

NO


Input

5 4
3 2 4 3 5


Output

YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int take() {
  int n;
  scanf("%d", &n);
  return n;
}
double ttake() {
  double n;
  scanf("%lf", &n);
  return n;
}
long long takes() {
  long long n;
  scanf("%lld", &n);
  return n;
}
int cas;
using namespace std;
bool approximatelyEqual(float a, float b, float epsilon) {
  return fabs(a - b) <= ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool essentiallyEqual(float a, float b, float epsilon) {
  return fabs(a - b) <= ((fabs(a) > fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool definitelyGreaterThan(float a, float b, float epsilon) {
  return (a - b) > ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
bool definitelyLessThan(float a, float b, float epsilon) {
  return (b - a) > ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}
vector<int> v1[105];
int main() {
  int tc = 1;
  while (tc--) {
    int n = take(), col = take(), arr[n + 1];
    for (int i = 1; i <= n; i += 1) arr[i] = take();
    for (int i = 1; i <= col; i += 1) {
      int mn = INT_MAX, nai = 0;
      for (int k = 1; k <= n; k += 1) {
        if (arr[k])
          mn = min(mn, arr[k]);
        else
          nai++;
      }
      if (mn == INT_MAX) break;
      if (nai) mn = 0;
      for (int k = 1; k <= n; k += 1) {
        if (arr[k]) {
          int d = mn;
          if (arr[k] - mn > 0) d++;
          for (int j = 1; j <= d; j += 1) v1[k].push_back(i);
          arr[k] -= d;
        }
      }
    }
    for (int i = 1; i <= n; i += 1) {
      if (arr[i]) {
        printf("NO");
        { return 0; };
      }
    }
    printf("YES\n");
    for (int i = 1; i <= n; i += 1) {
      for (int k = 0; k <= v1[i].size() - 1; k += 1) printf("%d ", v1[i][k]);
      printf("\n");
    }
  }
  { return 0; };
}