#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 + 10;
const int MOD = 1000000007;
inline int inc(int a, int b) {
  return (a + b >= MOD) ? (a + b - MOD) : (a + b);
}
inline int dec(int a, int b) { return (a >= b) ? (a - b) : (a + MOD - b); }
inline int mul(int a, int b) { return 1LL * a * b % MOD; }
inline int power(int x, int k) {
  if (k < 0) return 0;
  if (k == 0) return 1;
  int tmp = 1;
  while (k) {
    if (k & 1) tmp = mul(tmp, x);
    x = mul(x, x);
    k >>= 1;
  }
  return tmp;
}
int fac[maxn], ifac[maxn];
inline void prepare(int N) {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = mul(fac[i - 1], i);
  ifac[N] = power(fac[N], MOD - 2);
  for (int i = N - 1; i >= 1; i--) ifac[i] = mul(ifac[i + 1], i + 1);
}
inline int C(int n, int m) {
  if (n < 0 || m < 0 || n < m) return 0;
  return mul(fac[n], mul(ifac[m], ifac[n - m]));
}
int n, k, a[maxn];
int main() {
  prepare(200000);
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int p = 0;
  for (int x = 0; x <= k - 1; x++) {
    int tmp = (x & 1) ? (MOD - 1) : (1);
    tmp = mul(tmp, ifac[x]);
    tmp = mul(tmp, ifac[k - 1 - x]);
    int now = power(k - x, n - 1);
    now = inc(now, mul(n - 1, power(k - x, n - 2)));
    tmp = mul(tmp, now);
    p = inc(p, tmp);
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) sum = inc(sum, a[i]);
  int Ans = mul(sum, p);
  printf("%d\n", Ans);
  return 0;
}

### Prompt

Please formulate 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 = 200000 + 10;
const int MOD = 1000000007;
inline int inc(int a, int b) {
  return (a + b >= MOD) ? (a + b - MOD) : (a + b);
}
inline int dec(int a, int b) { return (a >= b) ? (a - b) : (a + MOD - b); }
inline int mul(int a, int b) { return 1LL * a * b % MOD; }
inline int power(int x, int k) {
  if (k < 0) return 0;
  if (k == 0) return 1;
  int tmp = 1;
  while (k) {
    if (k & 1) tmp = mul(tmp, x);
    x = mul(x, x);
    k >>= 1;
  }
  return tmp;
}
int fac[maxn], ifac[maxn];
inline void prepare(int N) {
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = mul(fac[i - 1], i);
  ifac[N] = power(fac[N], MOD - 2);
  for (int i = N - 1; i >= 1; i--) ifac[i] = mul(ifac[i + 1], i + 1);
}
inline int C(int n, int m) {
  if (n < 0 || m < 0 || n < m) return 0;
  return mul(fac[n], mul(ifac[m], ifac[n - m]));
}
int n, k, a[maxn];
int main() {
  prepare(200000);
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int p = 0;
  for (int x = 0; x <= k - 1; x++) {
    int tmp = (x & 1) ? (MOD - 1) : (1);
    tmp = mul(tmp, ifac[x]);
    tmp = mul(tmp, ifac[k - 1 - x]);
    int now = power(k - x, n - 1);
    now = inc(now, mul(n - 1, power(k - x, n - 2)));
    tmp = mul(tmp, now);
    p = inc(p, tmp);
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) sum = inc(sum, a[i]);
  int Ans = mul(sum, p);
  printf("%d\n", Ans);
  return 0;
}

```