#include <bits/stdc++.h>
using namespace std;
const long long N = 200 * 1001, MOD = 1e9 + 7;
long long n, k, a[N], fac[N], inv[N], sum;
long long _sum(long long a, long long b) { return (a + b) % MOD; }
long long _sub(long long a, long long b) { return (a - b + MOD) % MOD; }
long long _mul(long long a, long long b) {
  return (1LL * (a % MOD) * (b % MOD)) % MOD;
}
long long power(long long a, long long b) {
  if (!b) return 1LL;
  long long t = power(a, b / 2LL);
  if (b % 2) return _mul(_mul(t, t), a);
  return _mul(t, t);
}
void pre() {
  fac[0] = 1;
  for (long long i = 1; i < N; i++) fac[i] = _mul(fac[i - 1], i);
  inv[N - 1] = power(fac[N - 1], MOD - 2);
  for (long long i = N - 2; 0 <= i; i--) inv[i] = _mul(inv[i + 1], i + 1);
}
long long C(long long n, long long k) {
  if (n < k) return 0LL;
  if (!k || n == k) return 1LL;
  return _mul(_mul(fac[n], inv[k]), inv[n - k]);
}
long long stirling(long long n, long long k) {
  long long res = 0;
  for (long long i = 0; i <= k; i++) {
    long long tt = _mul(power(k - i, n), C(k, i));
    if (i % 2)
      res = _sub(res, tt);
    else
      res = _sum(res, tt);
  }
  return _mul(res, inv[k]);
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  pre();
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i] %= MOD;
    sum = _sum(sum, a[i]);
  }
  if (k == 1) {
    cout << _mul(n, sum) << "\n";
    return 0;
  }
  if (n == k) {
    cout << sum << "\n";
    return 0;
  }
  long long t = stirling(n - 1, k - 1);
  long long t2 = stirling(n - 1, k);
  t2 = _mul(t2, n - 1 + k);
  cout << _mul(sum, _sum(t, t2)) << "\n";
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

You are given a set of n elements indexed from 1 to n. The weight of i-th element is wi. The weight of some subset of a given set is denoted as <image>. The weight of some partition R of a given set into k subsets is <image> (recall that a partition of a given set is a set of its subsets such that every element of the given set belongs to exactly one subset in partition).

Calculate the sum of weights of all partitions of a given set into exactly k non-empty subsets, and print it modulo 109 + 7. Two partitions are considered different iff there exist two elements x and y such that they belong to the same set in one of the partitions, and to different sets in another partition.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2·105) — the number of elements and the number of subsets in each partition, respectively.

The second line contains n integers wi (1 ≤ wi ≤ 109)— weights of elements of the set.

Output

Print one integer — the sum of weights of all partitions of a given set into k non-empty subsets, taken modulo 109 + 7.

Examples

Input

4 2
2 3 2 3


Output

160


Input

5 2
1 2 3 4 5


Output

645

Note

Possible partitions in the first sample:

  1. {{1, 2, 3}, {4}}, W(R) = 3·(w1 + w2 + w3) + 1·w4 = 24; 
  2. {{1, 2, 4}, {3}}, W(R) = 26; 
  3. {{1, 3, 4}, {2}}, W(R) = 24; 
  4. {{1, 2}, {3, 4}}, W(R) = 2·(w1 + w2) + 2·(w3 + w4) = 20; 
  5. {{1, 3}, {2, 4}}, W(R) = 20; 
  6. {{1, 4}, {2, 3}}, W(R) = 20; 
  7. {{1}, {2, 3, 4}}, W(R) = 26; 



Possible partitions in the second sample:

  1. {{1, 2, 3, 4}, {5}}, W(R) = 45; 
  2. {{1, 2, 3, 5}, {4}}, W(R) = 48; 
  3. {{1, 2, 4, 5}, {3}}, W(R) = 51; 
  4. {{1, 3, 4, 5}, {2}}, W(R) = 54; 
  5. {{2, 3, 4, 5}, {1}}, W(R) = 57; 
  6. {{1, 2, 3}, {4, 5}}, W(R) = 36; 
  7. {{1, 2, 4}, {3, 5}}, W(R) = 37; 
  8. {{1, 2, 5}, {3, 4}}, W(R) = 38; 
  9. {{1, 3, 4}, {2, 5}}, W(R) = 38; 
  10. {{1, 3, 5}, {2, 4}}, W(R) = 39; 
  11. {{1, 4, 5}, {2, 3}}, W(R) = 40; 
  12. {{2, 3, 4}, {1, 5}}, W(R) = 39; 
  13. {{2, 3, 5}, {1, 4}}, W(R) = 40; 
  14. {{2, 4, 5}, {1, 3}}, W(R) = 41; 
  15. {{3, 4, 5}, {1, 2}}, W(R) = 42. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 200 * 1001, MOD = 1e9 + 7;
long long n, k, a[N], fac[N], inv[N], sum;
long long _sum(long long a, long long b) { return (a + b) % MOD; }
long long _sub(long long a, long long b) { return (a - b + MOD) % MOD; }
long long _mul(long long a, long long b) {
  return (1LL * (a % MOD) * (b % MOD)) % MOD;
}
long long power(long long a, long long b) {
  if (!b) return 1LL;
  long long t = power(a, b / 2LL);
  if (b % 2) return _mul(_mul(t, t), a);
  return _mul(t, t);
}
void pre() {
  fac[0] = 1;
  for (long long i = 1; i < N; i++) fac[i] = _mul(fac[i - 1], i);
  inv[N - 1] = power(fac[N - 1], MOD - 2);
  for (long long i = N - 2; 0 <= i; i--) inv[i] = _mul(inv[i + 1], i + 1);
}
long long C(long long n, long long k) {
  if (n < k) return 0LL;
  if (!k || n == k) return 1LL;
  return _mul(_mul(fac[n], inv[k]), inv[n - k]);
}
long long stirling(long long n, long long k) {
  long long res = 0;
  for (long long i = 0; i <= k; i++) {
    long long tt = _mul(power(k - i, n), C(k, i));
    if (i % 2)
      res = _sub(res, tt);
    else
      res = _sum(res, tt);
  }
  return _mul(res, inv[k]);
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  pre();
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i] %= MOD;
    sum = _sum(sum, a[i]);
  }
  if (k == 1) {
    cout << _mul(n, sum) << "\n";
    return 0;
  }
  if (n == k) {
    cout << sum << "\n";
    return 0;
  }
  long long t = stirling(n - 1, k - 1);
  long long t2 = stirling(n - 1, k);
  t2 = _mul(t2, n - 1 + k);
  cout << _mul(sum, _sum(t, t2)) << "\n";
}

```