#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int n, k, w[200010], fac[200010] = {1}, inv[200010] = {1};
int modpow(int a, long long n, int mod = MOD) {
  int r = 1, b = a;
  while (n > 0) {
    if (n & 1) r = 1ll * r * b % mod;
    b = 1ll * b * b % mod;
    n >>= 1;
  }
  return r;
}
long long nCk(int n, int k) {
  if (k < 0 || k > n) return 0;
  return 1ll * fac[n] * inv[k] % MOD * inv[n - k] % MOD;
}
long long nSk(int n, int k) {
  if (n == 0 && k == 0) return 1;
  if (k < 0 || k > n) return 0;
  long long res = 0;
  for (int i = 0; i <= k; i++) {
    int v = (k + i) % 2 ? -1 : 1;
    res = (res + v * nCk(k, i) * modpow(i, n) % MOD + MOD) % MOD;
  }
  res = res * inv[k] % MOD;
  return res;
}
int main() {
  for (int i = 1; i <= 200000; i++) fac[i] = 1ll * i * fac[i - 1] % MOD;
  for (int i = 1; i <= 200000; i++) inv[i] = modpow(fac[i], MOD - 2);
  scanf("%d%d", &n, &k);
  long long sum = 0;
  for (int i = 1; i <= n; i++) scanf("%d", w + i), sum += w[i];
  sum %= MOD;
  printf("%lld", (nSk(n, k) + (n - 1) * nSk(n - 1, k)) % MOD * sum % MOD);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 int MOD = 1e9 + 7;
int n, k, w[200010], fac[200010] = {1}, inv[200010] = {1};
int modpow(int a, long long n, int mod = MOD) {
  int r = 1, b = a;
  while (n > 0) {
    if (n & 1) r = 1ll * r * b % mod;
    b = 1ll * b * b % mod;
    n >>= 1;
  }
  return r;
}
long long nCk(int n, int k) {
  if (k < 0 || k > n) return 0;
  return 1ll * fac[n] * inv[k] % MOD * inv[n - k] % MOD;
}
long long nSk(int n, int k) {
  if (n == 0 && k == 0) return 1;
  if (k < 0 || k > n) return 0;
  long long res = 0;
  for (int i = 0; i <= k; i++) {
    int v = (k + i) % 2 ? -1 : 1;
    res = (res + v * nCk(k, i) * modpow(i, n) % MOD + MOD) % MOD;
  }
  res = res * inv[k] % MOD;
  return res;
}
int main() {
  for (int i = 1; i <= 200000; i++) fac[i] = 1ll * i * fac[i - 1] % MOD;
  for (int i = 1; i <= 200000; i++) inv[i] = modpow(fac[i], MOD - 2);
  scanf("%d%d", &n, &k);
  long long sum = 0;
  for (int i = 1; i <= n; i++) scanf("%d", w + i), sum += w[i];
  sum %= MOD;
  printf("%lld", (nSk(n, k) + (n - 1) * nSk(n - 1, k)) % MOD * sum % MOD);
  return 0;
}

```