#include <bits/stdc++.h>
using namespace std;
const int INF = 1001001001;
const long long INFLL = 1001001001001001001LL;
template <typename T>
void pv(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
template <typename T>
void chmin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
void chmax(T& a, T b) {
  if (a < b) a = b;
}
int in() {
  int x;
  scanf("%d", &x);
  return x;
}
double fin() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const long long MO = 1000000007;
const int MAXN = 200050;
long long fac[MAXN], inv[MAXN], ifac[MAXN];
long long comb(int n, int r) {
  return fac[n] * ifac[r] % MO * ifac[n - r] % MO;
}
long long powmod(long long x, long long e) {
  long long v = 1;
  for (; e; e >>= 1, (x *= x) %= MO)
    if (e & 1) (v *= x) %= MO;
  return v;
}
int main() {
  inv[1] = 1;
  for (int i = 2; i < MAXN; ++i)
    inv[i] = (MO - (MO / i) * inv[MO % i] % MO) % MO;
  fac[0] = ifac[0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    fac[i] = fac[i - 1] * i % MO;
    ifac[i] = ifac[i - 1] * inv[i] % MO;
  }
  const int N = in();
  const int K = in();
  if (N == 1) {
    printf("%d\n", in());
    return 0;
  }
  long long sum = 0;
  for (int j = 0; j < K; ++j) {
    const long long x = powmod(1 + j, N - 1);
    const long long y = powmod(1 + j, N - 2) * (N - 1) % MO;
    const long long v = comb(K - 1, j) * (x + y) % MO;
    (sum += (K - j - 1) % 2 ? (MO - v) : v) %= MO;
  }
  long long w = 0;
  for (int i = 0; i < N; ++i) {
    (w += in()) %= MO;
  }
  printf("%d\n", (int)(sum * w % MO * ifac[K - 1] % MO));
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 INF = 1001001001;
const long long INFLL = 1001001001001001001LL;
template <typename T>
void pv(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
template <typename T>
void chmin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
void chmax(T& a, T b) {
  if (a < b) a = b;
}
int in() {
  int x;
  scanf("%d", &x);
  return x;
}
double fin() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
const long long MO = 1000000007;
const int MAXN = 200050;
long long fac[MAXN], inv[MAXN], ifac[MAXN];
long long comb(int n, int r) {
  return fac[n] * ifac[r] % MO * ifac[n - r] % MO;
}
long long powmod(long long x, long long e) {
  long long v = 1;
  for (; e; e >>= 1, (x *= x) %= MO)
    if (e & 1) (v *= x) %= MO;
  return v;
}
int main() {
  inv[1] = 1;
  for (int i = 2; i < MAXN; ++i)
    inv[i] = (MO - (MO / i) * inv[MO % i] % MO) % MO;
  fac[0] = ifac[0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    fac[i] = fac[i - 1] * i % MO;
    ifac[i] = ifac[i - 1] * inv[i] % MO;
  }
  const int N = in();
  const int K = in();
  if (N == 1) {
    printf("%d\n", in());
    return 0;
  }
  long long sum = 0;
  for (int j = 0; j < K; ++j) {
    const long long x = powmod(1 + j, N - 1);
    const long long y = powmod(1 + j, N - 2) * (N - 1) % MO;
    const long long v = comb(K - 1, j) * (x + y) % MO;
    (sum += (K - j - 1) % 2 ? (MO - v) : v) %= MO;
  }
  long long w = 0;
  for (int i = 0; i < N; ++i) {
    (w += in()) %= MO;
  }
  printf("%d\n", (int)(sum * w % MO * ifac[K - 1] % MO));
  return 0;
}

```