#include <bits/stdc++.h>
using namespace std;
mt19937_64 rnd;
const long long maxn = 2e5 + 10;
const long long mod = 1e9 + 7;
const long long base = 1e18;
long long gt[maxn];
long long gtv[maxn];
long long mu(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long t = mu(a, n / 2);
  if (n % 2 == 0) return (t * t) % mod;
  return ((t * t) % mod * a) % mod;
}
void setup() {
  gt[0] = 1;
  for (int i = 1; i < maxn; i++) {
    gt[i] = (gt[i - 1] * i) % mod;
  }
  gtv[maxn - 1] = mu(gt[maxn - 1], mod - 2);
  for (int i = maxn - 2; i >= 0; i--) {
    gtv[i] = (gtv[i + 1] * (i + 1)) % mod;
  }
}
long long nck(long long n, long long k) {
  if (n < k) return 0;
  return ((gt[n] * gtv[k]) % mod * gtv[n - k]) % mod;
}
long long get(long long n, long long k) {
  if (n < k) return 0;
  long long nw = mu(gt[k], mod - 2);
  long long res = 0;
  for (int i = 0; i <= k; i++) {
    long long t = (nck(k, i) * ((i % 2 ? -1 : 1) + mod)) % mod * mu(k - i, n);
    res = (res + t) % mod;
  }
  return (res * nw) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (fopen("t.inp", "r")) {
    freopen("test.inp", "r", stdin);
    freopen("test.out", "w", stdout);
  }
  setup();
  long long n, k;
  cin >> n >> k;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    ans = (ans + x) % mod;
  }
  cout << ((get(n, k) + (n - 1) * get(n - 1, k)) % mod * ans) % mod;
}

### Prompt

Develop a solution in cpp 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;
mt19937_64 rnd;
const long long maxn = 2e5 + 10;
const long long mod = 1e9 + 7;
const long long base = 1e18;
long long gt[maxn];
long long gtv[maxn];
long long mu(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long t = mu(a, n / 2);
  if (n % 2 == 0) return (t * t) % mod;
  return ((t * t) % mod * a) % mod;
}
void setup() {
  gt[0] = 1;
  for (int i = 1; i < maxn; i++) {
    gt[i] = (gt[i - 1] * i) % mod;
  }
  gtv[maxn - 1] = mu(gt[maxn - 1], mod - 2);
  for (int i = maxn - 2; i >= 0; i--) {
    gtv[i] = (gtv[i + 1] * (i + 1)) % mod;
  }
}
long long nck(long long n, long long k) {
  if (n < k) return 0;
  return ((gt[n] * gtv[k]) % mod * gtv[n - k]) % mod;
}
long long get(long long n, long long k) {
  if (n < k) return 0;
  long long nw = mu(gt[k], mod - 2);
  long long res = 0;
  for (int i = 0; i <= k; i++) {
    long long t = (nck(k, i) * ((i % 2 ? -1 : 1) + mod)) % mod * mu(k - i, n);
    res = (res + t) % mod;
  }
  return (res * nw) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  if (fopen("t.inp", "r")) {
    freopen("test.inp", "r", stdin);
    freopen("test.out", "w", stdout);
  }
  setup();
  long long n, k;
  cin >> n >> k;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    ans = (ans + x) % mod;
  }
  cout << ((get(n, k) + (n - 1) * get(n - 1, k)) % mod * ans) % mod;
}

```