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;
inline int inp() { return 0; }
inline int inp(bool& A) {
  if (cin >> A) return 1;
  return -1;
}
inline int inp(char& A) { return scanf(" %c", &A); }
inline int inp(int& A) { return scanf("%d", &A); }
inline int inp(float& A) {
  if (cin >> A) return 1;
  return -1;
}
inline int inp(long long& A) { return scanf("%I64d", &A); }
inline int inp(double& A) { return scanf("%lf", &A); }
inline int inp(char* A) { return scanf("%s", A); }
inline int inp(string& A) {
  if (cin >> A) return 1;
  return -1;
}
template <class Front, class... Queue>
inline int inp(Front& A, Queue&... B) {
  return inp(A) + inp(B...);
}
inline int oup() { return 0; }
inline int oup(bool A) {
  if (cout << A) return 1;
  return -1;
}
inline int oup(char A) { return printf("%c", A); }
inline int oup(int A) { return printf("%d", A); }
inline int oup(float A) {
  if (cout << A) return 1;
  return -1;
}
inline int oup(long long A) { return printf("%I64d", A); }
inline int oup(double A) { return printf("%lf", A); }
inline int oup(char* A) { return printf(A); }
inline int oup(const char* A) { return printf("%s", A); }
inline int oup(string& A) {
  if (cout << A) return 1;
  return -1;
}
template <class Front, class... Queue>
inline int oup(Front A, Queue... B) {
  return oup(A) + oup(B...);
}
template <class T>
inline void remax(T& A, T B) {
  if (A < B) A = B;
}
template <class T>
inline void remin(T& A, T B) {
  if (A > B) A = B;
}
string ToString(long long num) {
  string ret;
  do {
    ret += ((num % 10) + '0');
    num /= 10;
  } while (num);
  reverse(ret.begin(), ret.end());
  return ret;
}
long long ToNumber(string s) {
  long long r = 0, p = 1;
  for (int i = s.size() - 1; i >= 0; --i) r += (s[i] - '0') * p, p *= 10;
  return r;
}
long long Gcd(long long a, long long b) {
  while (a %= b ^= a ^= b ^= a)
    ;
  return b;
}
long long Power(long long base, long long power) {
  long long ret = 1;
  while (power) {
    if (power & 1) ret *= base;
    power >>= 1;
    base *= base;
  }
  return ret;
}
int Log(long long num, long long base) {
  int ret = 0;
  while (num) {
    ++ret;
    num /= base;
  }
  return ret;
}
int Count(long long mask) {
  int ret = 0;
  while (mask) {
    if (mask & 1) ++ret;
    mask >>= 1;
  }
  return ret;
}
inline void run() {
in:;
  int n, k;
  if (inp(n, k) != 2) return;
  int arr[100];
  for (int i = 0; i < n;) inp(arr[i++]);
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j)
      if (abs(arr[i] - arr[j]) > k) {
        oup("NO\n");
        goto in;
      }
  oup("YES\n");
  for (int i = 0; i < n; ++i) {
    int tmp = 1;
    while (arr[i]--) {
      oup(tmp++);
      if (arr[i]) oup(' ');
      if (tmp == k + 1) tmp = 1;
    }
    oup('\n');
  }
  goto in;
}
int main() {
  run();
  return 0;
}