#include <bits/stdc++.h>
using namespace std;
const int N = 200200;
const int Mod = 1e9 + 7;
int Inv[N];
int Fact[N];
int iFact[N];
void Add_self(int& x, int y) {
  if ((x += y) >= Mod) x -= Mod;
}
int Add(int x, int y) { return Add_self(x, y), x; }
int Mul(int x, int y) { return (long long)x * y % Mod; }
int Fp(int x, int y) {
  if (!y) return 1;
  if (y & 1) return Mul(x, Fp(x, y - 1));
  return Fp(Mul(x, x), y >> 1);
}
int C(int n, int k) {
  if (n < 0 || k < 0 || n < k) return 0;
  return Mul(Fact[n], Mul(iFact[k], iFact[n - k]));
}
int StirlingNumber2(int n, int k) {
  if (n < k) return 0;
  if (k == 1 || k == n) return 1;
  int Res = 0;
  for (int i = 0; i <= k; i++) {
    int Staff = Mul(C(k, i), Fp(i, n));
    Staff = Mul(Staff, Fp(Mod - 1, k - i));
    Add_self(Res, Staff);
  }
  Res = Mul(Res, iFact[k]);
  return Res;
}
int main() {
  Inv[1] = 1;
  for (int i = 2; i < N; i++) Inv[i] = (Mod - Mul(Mod / i, Inv[Mod % i])) % Mod;
  Fact[0] = iFact[0] = 1;
  for (int i = 1; i < N; i++) Fact[i] = Mul(i, Fact[i - 1]);
  for (int i = 1; i < N; i++) iFact[i] = Mul(Inv[i], iFact[i - 1]);
  int n, k;
  cin >> n >> k;
  int Sum = 0;
  for (int i = 0, x; i < n; i++) scanf("%d", &x), Add_self(Sum, x);
  int Coff = Add(StirlingNumber2(n, k), Mul(n - 1, StirlingNumber2(n - 1, k)));
  int Ans = Mul(Sum, Coff);
  cout << Ans << endl;
}

### Prompt

In CPP, your task is to solve 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 int N = 200200;
const int Mod = 1e9 + 7;
int Inv[N];
int Fact[N];
int iFact[N];
void Add_self(int& x, int y) {
  if ((x += y) >= Mod) x -= Mod;
}
int Add(int x, int y) { return Add_self(x, y), x; }
int Mul(int x, int y) { return (long long)x * y % Mod; }
int Fp(int x, int y) {
  if (!y) return 1;
  if (y & 1) return Mul(x, Fp(x, y - 1));
  return Fp(Mul(x, x), y >> 1);
}
int C(int n, int k) {
  if (n < 0 || k < 0 || n < k) return 0;
  return Mul(Fact[n], Mul(iFact[k], iFact[n - k]));
}
int StirlingNumber2(int n, int k) {
  if (n < k) return 0;
  if (k == 1 || k == n) return 1;
  int Res = 0;
  for (int i = 0; i <= k; i++) {
    int Staff = Mul(C(k, i), Fp(i, n));
    Staff = Mul(Staff, Fp(Mod - 1, k - i));
    Add_self(Res, Staff);
  }
  Res = Mul(Res, iFact[k]);
  return Res;
}
int main() {
  Inv[1] = 1;
  for (int i = 2; i < N; i++) Inv[i] = (Mod - Mul(Mod / i, Inv[Mod % i])) % Mod;
  Fact[0] = iFact[0] = 1;
  for (int i = 1; i < N; i++) Fact[i] = Mul(i, Fact[i - 1]);
  for (int i = 1; i < N; i++) iFact[i] = Mul(Inv[i], iFact[i - 1]);
  int n, k;
  cin >> n >> k;
  int Sum = 0;
  for (int i = 0, x; i < n; i++) scanf("%d", &x), Add_self(Sum, x);
  int Coff = Add(StirlingNumber2(n, k), Mul(n - 1, StirlingNumber2(n - 1, k)));
  int Ans = Mul(Sum, Coff);
  cout << Ans << endl;
}

```