#include <bits/stdc++.h>
using namespace std;
const int maxn = 200001;
const int lbt = 1e9 + 7;
int n, k;
int rev[maxn], fac[maxn], rfac[maxn];
int prime[maxn], cnt;
bool not_p[maxn];
int powa[maxn], powb[maxn];
int xs[maxn];
int w[maxn];
int comb(int n, int m);
int fast_pow(int a, int t);
int main() {
  scanf("%d%d", &n, &k);
  rev[1] = 1;
  for (int i = 2; i <= k; i++) {
    rev[i] = (long long)(lbt - lbt / i) * rev[lbt % i] % lbt;
  }
  fac[0] = rfac[0] = 1;
  for (int i = 1; i <= k; i++) {
    fac[i] = (long long)i * fac[i - 1] % lbt;
    rfac[i] = (long long)rev[i] * rfac[i - 1] % lbt;
  }
  not_p[1] = 1;
  powa[1] = powb[1] = 1;
  for (int i = 2; i <= k; i++) {
    if (!not_p[i]) {
      prime[++cnt] = i;
      powa[i] = fast_pow(i, n);
      powb[i] = fast_pow(i, n - 1);
    }
    for (int j = 1; j <= cnt && i * prime[j] <= k; j++) {
      not_p[i * prime[j]] = 1;
      powa[i * prime[j]] = (long long)powa[i] * powa[prime[j]] % lbt;
      powb[i * prime[j]] = (long long)powb[i] * powb[prime[j]] % lbt;
      if (i % prime[j] == 0) break;
    }
  }
  xs[0] = 1;
  for (int i = 1; i <= k; i++) {
    xs[i] = lbt - xs[i - 1];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
  }
  if (n == 1) {
    printf("%d", w[1]);
    return 0;
  }
  int a = 0;
  int b = 0;
  for (int i = 0; i < k; i++) {
    a = (a + (long long)xs[i] * comb(k, i) % lbt * powa[k - i]) % lbt;
    b = (b + (long long)xs[i] * comb(k, i) % lbt * powb[k - i]) % lbt;
  }
  a = (long long)rfac[k] * a % lbt;
  b = (long long)rfac[k] * b % lbt;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = (ans + (long long)w[i] * a) % lbt;
    ans = (ans + (long long)w[i] * (n - 1) % lbt * b) % lbt;
  }
  printf("%d", ans);
  return 0;
}
int comb(int n, int m) {
  if (n < 0 || m < 0 || m > n) return 0;
  return (long long)fac[n] * rfac[m] % lbt * rfac[n - m] % lbt;
}
int fast_pow(int a, int t) {
  int ans = 1;
  for (int i = 31; i >= 0; i--) {
    ans = (long long)ans * ans % lbt;
    if (t >> i & 1) ans = (long long)a * ans % lbt;
  }
  return ans;
}

### Prompt

Your task is to create 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 maxn = 200001;
const int lbt = 1e9 + 7;
int n, k;
int rev[maxn], fac[maxn], rfac[maxn];
int prime[maxn], cnt;
bool not_p[maxn];
int powa[maxn], powb[maxn];
int xs[maxn];
int w[maxn];
int comb(int n, int m);
int fast_pow(int a, int t);
int main() {
  scanf("%d%d", &n, &k);
  rev[1] = 1;
  for (int i = 2; i <= k; i++) {
    rev[i] = (long long)(lbt - lbt / i) * rev[lbt % i] % lbt;
  }
  fac[0] = rfac[0] = 1;
  for (int i = 1; i <= k; i++) {
    fac[i] = (long long)i * fac[i - 1] % lbt;
    rfac[i] = (long long)rev[i] * rfac[i - 1] % lbt;
  }
  not_p[1] = 1;
  powa[1] = powb[1] = 1;
  for (int i = 2; i <= k; i++) {
    if (!not_p[i]) {
      prime[++cnt] = i;
      powa[i] = fast_pow(i, n);
      powb[i] = fast_pow(i, n - 1);
    }
    for (int j = 1; j <= cnt && i * prime[j] <= k; j++) {
      not_p[i * prime[j]] = 1;
      powa[i * prime[j]] = (long long)powa[i] * powa[prime[j]] % lbt;
      powb[i * prime[j]] = (long long)powb[i] * powb[prime[j]] % lbt;
      if (i % prime[j] == 0) break;
    }
  }
  xs[0] = 1;
  for (int i = 1; i <= k; i++) {
    xs[i] = lbt - xs[i - 1];
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
  }
  if (n == 1) {
    printf("%d", w[1]);
    return 0;
  }
  int a = 0;
  int b = 0;
  for (int i = 0; i < k; i++) {
    a = (a + (long long)xs[i] * comb(k, i) % lbt * powa[k - i]) % lbt;
    b = (b + (long long)xs[i] * comb(k, i) % lbt * powb[k - i]) % lbt;
  }
  a = (long long)rfac[k] * a % lbt;
  b = (long long)rfac[k] * b % lbt;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = (ans + (long long)w[i] * a) % lbt;
    ans = (ans + (long long)w[i] * (n - 1) % lbt * b) % lbt;
  }
  printf("%d", ans);
  return 0;
}
int comb(int n, int m) {
  if (n < 0 || m < 0 || m > n) return 0;
  return (long long)fac[n] * rfac[m] % lbt * rfac[n - m] % lbt;
}
int fast_pow(int a, int t) {
  int ans = 1;
  for (int i = 31; i >= 0; i--) {
    ans = (long long)ans * ans % lbt;
    if (t >> i & 1) ans = (long long)a * ans % lbt;
  }
  return ans;
}

```