#include <bits/stdc++.h>
using namespace std;
long long N, K, fact[200001], revFact[200001], mod = 1e9 + 7, w[200001];
inline long long mul(long long a, long long b) { return a * b % mod; }
long long powmod(long long a, long long x) {
  if (x == 0) return 1;
  if (x & 1) return mul(a, powmod(a, x - 1));
  long long temp = powmod(a, x >> 1);
  return mul(temp, temp);
}
void preprocess() {
  fact[0] = 1;
  for (int i = 1; i <= 200001 - 1; ++i) fact[i] = mul(fact[i - 1], i);
  revFact[200001 - 1] = powmod(fact[200001 - 1], mod - 2);
  for (int i = 200001 - 2; i >= 0; --i) revFact[i] = mul(revFact[i + 1], i + 1);
}
inline long long comb(long long x, long long y) {
  return mul(fact[y], mul(revFact[x], revFact[y - x]));
}
long long stirling(long long k, long long n) {
  if (k > n) return 0;
  long long res = 0;
  long long temp = (k % 2 == 0) ? 1 : -1;
  for (int j = 0; j <= k; ++j) {
    res += temp * mul(comb(j, k), powmod(j, n));
    res = (res + mod) % mod;
    temp *= -1;
  }
  return mul(res, revFact[k]);
}
void setup() {
  preprocess();
  cin >> N >> K;
  for (int i = 1; i <= N; ++i) cin >> w[i];
}
void process() {
  long long t1 = stirling(K, N), t2 = stirling(K, N - 1);
  long long res = 0;
  long long sum = 0;
  for (int i = 1; i <= N; ++i) sum = (sum + w[i]) % mod;
  for (int i = 1; i <= N; ++i) {
    long long temp = (sum - w[i] + mod) % mod;
    res = (res + mul(w[i], t1) + mul(temp, t2)) % mod;
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  setup();
  process();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long N, K, fact[200001], revFact[200001], mod = 1e9 + 7, w[200001];
inline long long mul(long long a, long long b) { return a * b % mod; }
long long powmod(long long a, long long x) {
  if (x == 0) return 1;
  if (x & 1) return mul(a, powmod(a, x - 1));
  long long temp = powmod(a, x >> 1);
  return mul(temp, temp);
}
void preprocess() {
  fact[0] = 1;
  for (int i = 1; i <= 200001 - 1; ++i) fact[i] = mul(fact[i - 1], i);
  revFact[200001 - 1] = powmod(fact[200001 - 1], mod - 2);
  for (int i = 200001 - 2; i >= 0; --i) revFact[i] = mul(revFact[i + 1], i + 1);
}
inline long long comb(long long x, long long y) {
  return mul(fact[y], mul(revFact[x], revFact[y - x]));
}
long long stirling(long long k, long long n) {
  if (k > n) return 0;
  long long res = 0;
  long long temp = (k % 2 == 0) ? 1 : -1;
  for (int j = 0; j <= k; ++j) {
    res += temp * mul(comb(j, k), powmod(j, n));
    res = (res + mod) % mod;
    temp *= -1;
  }
  return mul(res, revFact[k]);
}
void setup() {
  preprocess();
  cin >> N >> K;
  for (int i = 1; i <= N; ++i) cin >> w[i];
}
void process() {
  long long t1 = stirling(K, N), t2 = stirling(K, N - 1);
  long long res = 0;
  long long sum = 0;
  for (int i = 1; i <= N; ++i) sum = (sum + w[i]) % mod;
  for (int i = 1; i <= N; ++i) {
    long long temp = (sum - w[i] + mod) % mod;
    res = (res + mul(w[i], t1) + mul(temp, t2)) % mod;
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  setup();
  process();
  return 0;
}

```