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>
using namespace std;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void seive(vector<bool> &arr) {
  for (int64_t i = 2; i <= sqrt(1000001); i++) {
    if (arr[i] == 1) {
      for (int j = i * i; j <= 1000001; j += i) {
        arr[j] = 0;
      }
    }
  }
  arr[0] = 0;
  arr[1] = 0;
}
int binary_search(vector<int64_t> arr, int n) {
  int low = 1, high = arr.size() - 1, mid, leftGreater = arr.size() - 1;
  while (low <= high && low >= 0) {
    mid = low + (high - low) / 2;
    if (arr[mid] >= n) {
      leftGreater = mid;
      high = mid - 1;
    } else
      low = mid + 1;
  }
  return leftGreater;
}
int64_t dfs(vector<int64_t> arr, vector<int64_t> &visited,
            vector<int64_t> &height, int start) {
  visited[start] = 1;
  if (arr[start] == -1) {
    height[start] = 0;
    return 0;
  } else {
    if (visited[arr[start]] == 1) {
      height[start] = height[arr[start]] + 1;
      return height[start];
    } else {
      height[start] = dfs(arr, visited, height, arr[start]) + 1;
      return height[start];
    }
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> arr(n);
  vector<vector<int> > colors(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  int ctr = 0;
  while (true) {
    int mini = 1000;
    if (ctr == 0) {
      for (int i = 0; i < n; i++) {
        if (arr[i] > 0) mini = min(mini, arr[i]);
      }
    } else
      mini = 1;
    for (int i = 0; i < n; i++) {
      if (arr[i] > 0) {
        int temp = mini;
        if (arr[i] > mini and ctr == 0) temp = mini + 1;
        arr[i] -= temp;
        for (int j = 0; j < temp; j++) colors[i].push_back(ctr);
      }
    }
    int flag = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] > 0) flag = 1;
    }
    ctr += 1;
    if (flag == 0) break;
  }
  if (ctr <= k) {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < colors[i].size(); j++)
        cout << colors[i][j] + 1 << " ";
      cout << "\n";
    }
  } else {
    cout << "NO\n";
  }
  return 0;
}