#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 10;
const long long P = 1000 * 1000 * 1000 + 7;
long long n, k, sum, ans;
long long fac[N], a[N];
long long chert(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long ret = chert(x, y / 2);
  ret *= ret;
  ret %= P;
  if (y % 2) {
    ret *= x;
    ret %= P;
  }
  return ret;
}
long long ch(long long x, long long y) {
  if (x < y) {
    return 0;
  }
  long long ret = fac[x] * chert(fac[y], P - 2);
  ret %= P;
  ret *= chert(fac[x - y], P - 2);
  ret %= P;
  return ret;
}
long long f(long long n, long long k) {
  long long ret = 0;
  for (long long i = 1; i <= k; i++) {
    long long x = 1ll * ch(k, i) * chert(i, n);
    x %= P;
    if (i % 2 == k % 2)
      ret += x;
    else
      ret -= x - P;
  }
  ret %= P;
  return ret;
}
void set_fac() {
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i;
    fac[i] %= P;
  }
}
void read_input() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  sum %= P;
}
void solve() {
  set_fac();
  ans = f(n, k) + 1ll * f(n - 1, k) * (n - 1);
  ans %= P;
  ans *= sum;
  ans %= P;
  ans *= chert(fac[k], P - 2);
  ans %= P;
}
void write_output() { cout << ans << endl; }
int32_t main() {
  ios ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read_input(), solve(), write_output();
  return 0;
}

### Prompt

Generate 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 long long N = 2e5 + 10;
const long long P = 1000 * 1000 * 1000 + 7;
long long n, k, sum, ans;
long long fac[N], a[N];
long long chert(long long x, long long y) {
  if (y == 0) {
    return 1;
  }
  long long ret = chert(x, y / 2);
  ret *= ret;
  ret %= P;
  if (y % 2) {
    ret *= x;
    ret %= P;
  }
  return ret;
}
long long ch(long long x, long long y) {
  if (x < y) {
    return 0;
  }
  long long ret = fac[x] * chert(fac[y], P - 2);
  ret %= P;
  ret *= chert(fac[x - y], P - 2);
  ret %= P;
  return ret;
}
long long f(long long n, long long k) {
  long long ret = 0;
  for (long long i = 1; i <= k; i++) {
    long long x = 1ll * ch(k, i) * chert(i, n);
    x %= P;
    if (i % 2 == k % 2)
      ret += x;
    else
      ret -= x - P;
  }
  ret %= P;
  return ret;
}
void set_fac() {
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i;
    fac[i] %= P;
  }
}
void read_input() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  sum %= P;
}
void solve() {
  set_fac();
  ans = f(n, k) + 1ll * f(n - 1, k) * (n - 1);
  ans %= P;
  ans *= sum;
  ans %= P;
  ans *= chert(fac[k], P - 2);
  ans %= P;
}
void write_output() { cout << ans << endl; }
int32_t main() {
  ios ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read_input(), solve(), write_output();
  return 0;
}

```