#include <bits/stdc++.h>
using ll = long long;
const int mod = 1e9 + 7;
const int maxn = 1 << 20;
int n, k, w[maxn], sum, fac[maxn], ifac[maxn];
int qpow(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = (ll)res * a % mod;
    a = (ll)a * a % mod;
    b >>= 1;
  }
  return res;
}
int inv(int x) {
  x %= mod;
  return x == 1 ? 1 : (ll)(mod - mod / x) * inv(mod % x) % mod;
}
int binom(int n, int m) {
  if (n < m || n < 0 || m < 0) return 0;
  return (ll)fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  fac[0] = ifac[0] = ifac[1] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = (ll)fac[i - 1] * i % mod;
  for (int i = 2; i <= n; ++i)
    ifac[i] = (ll)(mod - mod / i) * ifac[mod % i] % mod;
  for (int i = 1; i <= n; ++i) ifac[i] = (ll)ifac[i - 1] * ifac[i] % mod;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &w[i]);
    sum = (sum + w[i]) % mod;
  }
  if (k == 1) {
    printf("%lld", (ll)n * sum % mod);
    return 0;
  }
  k--;
  int ans = 0, N = n - 1;
  auto det = [](int x) { return x & 1 ? mod - 1 : 1; };
  for (int i = 0; i <= k; ++i) {
    int coef = 0, latt = 0;
    if (i == 0) {
      coef = (ll)binom(k, i) % mod * det(k - i) % mod;
      for (int j = 0; j <= N; ++j)
        latt = (latt + (ll)binom(N, j) * (j + 1) % mod * qpow(i, N - j) % mod) %
               mod;
    } else {
      coef = (ll)qpow(i, N) * binom(k, i) % mod * det(k - i) % mod;
      int q = inv(i);
      latt =
          (qpow(q + 1, N) + (ll)qpow(q + 1, N - 1) * N % mod * q % mod) % mod;
    }
    ans = (ans + (ll)coef * latt % mod) % mod;
  }
  printf("%lld", (ll)ans * ifac[k] % mod * sum % mod);
  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 ll = long long;
const int mod = 1e9 + 7;
const int maxn = 1 << 20;
int n, k, w[maxn], sum, fac[maxn], ifac[maxn];
int qpow(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = (ll)res * a % mod;
    a = (ll)a * a % mod;
    b >>= 1;
  }
  return res;
}
int inv(int x) {
  x %= mod;
  return x == 1 ? 1 : (ll)(mod - mod / x) * inv(mod % x) % mod;
}
int binom(int n, int m) {
  if (n < m || n < 0 || m < 0) return 0;
  return (ll)fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  fac[0] = ifac[0] = ifac[1] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = (ll)fac[i - 1] * i % mod;
  for (int i = 2; i <= n; ++i)
    ifac[i] = (ll)(mod - mod / i) * ifac[mod % i] % mod;
  for (int i = 1; i <= n; ++i) ifac[i] = (ll)ifac[i - 1] * ifac[i] % mod;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &w[i]);
    sum = (sum + w[i]) % mod;
  }
  if (k == 1) {
    printf("%lld", (ll)n * sum % mod);
    return 0;
  }
  k--;
  int ans = 0, N = n - 1;
  auto det = [](int x) { return x & 1 ? mod - 1 : 1; };
  for (int i = 0; i <= k; ++i) {
    int coef = 0, latt = 0;
    if (i == 0) {
      coef = (ll)binom(k, i) % mod * det(k - i) % mod;
      for (int j = 0; j <= N; ++j)
        latt = (latt + (ll)binom(N, j) * (j + 1) % mod * qpow(i, N - j) % mod) %
               mod;
    } else {
      coef = (ll)qpow(i, N) * binom(k, i) % mod * det(k - i) % mod;
      int q = inv(i);
      latt =
          (qpow(q + 1, N) + (ll)qpow(q + 1, N - 1) * N % mod * q % mod) % mod;
    }
    ans = (ans + (ll)coef * latt % mod) % mod;
  }
  printf("%lld", (ll)ans * ifac[k] % mod * sum % mod);
  return 0;
}

```