#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int add(int x, int y) { return x + y < mod ? x + y : x + y - mod; }
void ade(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int qpow(int x, int k) {
  int r = 1;
  while (k) {
    if (k & 1) r = 1ll * r * x % mod;
    k >>= 1;
    x = 1ll * x * x % mod;
  }
  return r;
}
int fac[500500], ifac[500500];
void init() {
  int ma = 5e5;
  fac[0] = 1;
  for (int i = 1; i <= ma; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[ma] = qpow(fac[ma], mod - 2);
  for (int i = ma - 1; i >= 0; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}
int C(int x, int y) { return 1ll * fac[x] * ifac[y] % mod * ifac[x - y] % mod; }
int S(int a, int b) {
  int res = 0;
  for (int i = 0; i <= b; ++i)
    if (i & 1)
      ade(res, mod - 1ll * C(b, i) * qpow(b - i, a) % mod);
    else
      ade(res, 1ll * C(b, i) * qpow(b - i, a) % mod);
  return 1ll * res * ifac[b] % mod;
}
int n, m;
int main() {
  init();
  scanf("%d%d", &n, &m);
  int sum = 0, xx;
  for (int i = 1; i <= n; ++i) scanf("%d", &xx), ade(sum, xx);
  int ans = add(S(n, m), 1ll * (n - 1) * S(n - 1, m) % mod);
  ans = 1ll * ans * sum % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate a cpp solution to 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 mod = 1e9 + 7;
int add(int x, int y) { return x + y < mod ? x + y : x + y - mod; }
void ade(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int qpow(int x, int k) {
  int r = 1;
  while (k) {
    if (k & 1) r = 1ll * r * x % mod;
    k >>= 1;
    x = 1ll * x * x % mod;
  }
  return r;
}
int fac[500500], ifac[500500];
void init() {
  int ma = 5e5;
  fac[0] = 1;
  for (int i = 1; i <= ma; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
  ifac[ma] = qpow(fac[ma], mod - 2);
  for (int i = ma - 1; i >= 0; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
}
int C(int x, int y) { return 1ll * fac[x] * ifac[y] % mod * ifac[x - y] % mod; }
int S(int a, int b) {
  int res = 0;
  for (int i = 0; i <= b; ++i)
    if (i & 1)
      ade(res, mod - 1ll * C(b, i) * qpow(b - i, a) % mod);
    else
      ade(res, 1ll * C(b, i) * qpow(b - i, a) % mod);
  return 1ll * res * ifac[b] % mod;
}
int n, m;
int main() {
  init();
  scanf("%d%d", &n, &m);
  int sum = 0, xx;
  for (int i = 1; i <= n; ++i) scanf("%d", &xx), ade(sum, xx);
  int ans = add(S(n, m), 1ll * (n - 1) * S(n - 1, m) % mod);
  ans = 1ll * ans * sum % mod;
  printf("%d\n", ans);
  return 0;
}

```