#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 1;
const int MOD = 1e9 + 7;
int add(int a, int b) { return (a += b) >= MOD ? a - MOD : a; }
int sub(int a, int b) { return (a -= b) < 0 ? a + MOD : a; }
int mul(int a, int b) { return (long long)a * b % MOD; }
int power(int a, long long n) {
  int res = 1;
  while (n) {
    if (n & 1LL) res = mul(res, a);
    a = mul(a, a);
    n >>= 1LL;
  }
  return res;
}
int inv(int a) { return power(a, MOD - 2); }
int fact[N], infact[N], n, k, sumW = 0;
int nCr(int n, int r) {
  if (r > n) return 0;
  return mul(fact[n], mul(infact[n - r], infact[r]));
}
int strirlingSecondKind(int n, int r) {
  int ans = 0;
  for (int i = 0; i <= r; ++i) {
    if ((r - i) & 1)
      ans = sub(ans, mul(nCr(r, i), power(i, n)));
    else
      ans = add(ans, mul(nCr(r, i), power(i, n)));
  }
  ans = mul(ans, infact[r]);
  return ans;
}
void prepare() {
  fact[0] = infact[0] = 1;
  for (int i = 1; i < N; ++i) fact[i] = mul(fact[i - 1], i);
  infact[N - 1] = inv(fact[N - 1]);
  for (int i = N - 2; i >= 1; --i) infact[i] = mul(infact[i + 1], i + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  prepare();
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    int w;
    cin >> w;
    sumW = add(sumW, w);
  }
  long long ans =
      add(strirlingSecondKind(n, k), mul(n - 1, strirlingSecondKind(n - 1, k)));
  ans = mul(ans, sumW);
  cout << ans;
  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;
const int N = 2e5 + 1;
const int MOD = 1e9 + 7;
int add(int a, int b) { return (a += b) >= MOD ? a - MOD : a; }
int sub(int a, int b) { return (a -= b) < 0 ? a + MOD : a; }
int mul(int a, int b) { return (long long)a * b % MOD; }
int power(int a, long long n) {
  int res = 1;
  while (n) {
    if (n & 1LL) res = mul(res, a);
    a = mul(a, a);
    n >>= 1LL;
  }
  return res;
}
int inv(int a) { return power(a, MOD - 2); }
int fact[N], infact[N], n, k, sumW = 0;
int nCr(int n, int r) {
  if (r > n) return 0;
  return mul(fact[n], mul(infact[n - r], infact[r]));
}
int strirlingSecondKind(int n, int r) {
  int ans = 0;
  for (int i = 0; i <= r; ++i) {
    if ((r - i) & 1)
      ans = sub(ans, mul(nCr(r, i), power(i, n)));
    else
      ans = add(ans, mul(nCr(r, i), power(i, n)));
  }
  ans = mul(ans, infact[r]);
  return ans;
}
void prepare() {
  fact[0] = infact[0] = 1;
  for (int i = 1; i < N; ++i) fact[i] = mul(fact[i - 1], i);
  infact[N - 1] = inv(fact[N - 1]);
  for (int i = N - 2; i >= 1; --i) infact[i] = mul(infact[i + 1], i + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  prepare();
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    int w;
    cin >> w;
    sumW = add(sumW, w);
  }
  long long ans =
      add(strirlingSecondKind(n, k), mul(n - 1, strirlingSecondKind(n - 1, k)));
  ans = mul(ans, sumW);
  cout << ans;
  return 0;
}

```