#include <bits/stdc++.h>
using namespace ::std;
const long long maxn = 3e5 + 500;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 500;
long long poww(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b = (b / 2);
  }
  return ans;
}
long long fac[maxn];
long long rfac[maxn];
long long ent(long long n, long long k) {
  if (k < 0 || k > n) {
    return 0;
  }
  return ((((fac[n] * rfac[k]) % mod) * rfac[n - k]) % mod);
}
long long s(long long n, long long k) {
  long long ans = 0;
  for (long long i = k; i >= 1; i--) {
    long long z;
    if (i % 2 == k % 2) {
      z = 1;
    } else {
      z = -1;
    }
    ans += z * poww(i, n) * ent(k, i);
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  return ans;
}
int main() {
  fac[0] = 1;
  rfac[0] = 1;
  for (long long i = 1; i < maxn; i++) {
    fac[i] = (fac[i - 1] * i) % mod;
    rfac[i] = poww(fac[i], mod - 2);
  }
  long long n, k, sum = 0;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    long long v;
    cin >> v;
    sum += v;
  }
  sum %= mod;
  long long za = 1;
  for (long long i = 1; i < k; i++) {
    za = (za * i) % mod;
  }
  za = poww(za, mod - 2);
  n--;
  long long v1 = s(n, k - 1);
  long long v2 = (((s(n, k) * (n + k)) % mod) * poww(k, mod - 2)) % mod;
  long long w = (v1 + v2) % mod;
  cout << (((sum * w) % mod) * za) % mod;
}

### 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 long long maxn = 3e5 + 500;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 500;
long long poww(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    b = (b / 2);
  }
  return ans;
}
long long fac[maxn];
long long rfac[maxn];
long long ent(long long n, long long k) {
  if (k < 0 || k > n) {
    return 0;
  }
  return ((((fac[n] * rfac[k]) % mod) * rfac[n - k]) % mod);
}
long long s(long long n, long long k) {
  long long ans = 0;
  for (long long i = k; i >= 1; i--) {
    long long z;
    if (i % 2 == k % 2) {
      z = 1;
    } else {
      z = -1;
    }
    ans += z * poww(i, n) * ent(k, i);
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  return ans;
}
int main() {
  fac[0] = 1;
  rfac[0] = 1;
  for (long long i = 1; i < maxn; i++) {
    fac[i] = (fac[i - 1] * i) % mod;
    rfac[i] = poww(fac[i], mod - 2);
  }
  long long n, k, sum = 0;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    long long v;
    cin >> v;
    sum += v;
  }
  sum %= mod;
  long long za = 1;
  for (long long i = 1; i < k; i++) {
    za = (za * i) % mod;
  }
  za = poww(za, mod - 2);
  n--;
  long long v1 = s(n, k - 1);
  long long v2 = (((s(n, k) * (n + k)) % mod) * poww(k, mod - 2)) % mod;
  long long w = (v1 + v2) % mod;
  cout << (((sum * w) % mod) * za) % mod;
}

```