#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int out = 0, fh = 1;
  char jp = getchar();
  while ((jp > '9' || jp < '0') && jp != '-') jp = getchar();
  if (jp == '-') fh = -1, jp = getchar();
  while (jp >= '0' && jp <= '9') out = out * 10 + jp - '0', jp = getchar();
  return out * fh;
}
void print(int x) {
  if (x >= 10) print(x / 10);
  putchar('0' + x % 10);
}
void write(int x, char c) {
  if (x < 0) putchar('-'), x = -x;
  print(x);
  putchar(c);
}
const int P = 1e9 + 7;
int add(int a, int b) { return a + b >= P ? a + b - P : a + b; }
void inc(int &a, int b) { a = add(a, b); }
int mul(int a, int b) { return 1LL * a * b % P; }
int fpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
const int N = 2e5 + 10;
int fac[N], invfac[N];
int S(int n, int k) {
  if (n < k) return 0;
  int s = 0;
  for (int i = 0; i <= k; ++i) {
    int t = mul(mul(invfac[k - i], invfac[i]), fpow(k - i, n));
    if (i & 1)
      inc(s, P - t);
    else
      inc(s, t);
  }
  return s;
}
int n, k, ans = 0;
int main() {
  n = read(), k = read();
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) inc(ans, read()), fac[i] = mul(fac[i - 1], i);
  invfac[n] = fpow(fac[n], P - 2);
  for (int i = n - 1; i >= 0; --i) invfac[i] = mul(invfac[i + 1], i + 1);
  int coef = add(mul(S(n - 1, k), n - 1), S(n, k));
  ans = mul(ans, coef);
  cout << ans << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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;
inline int read() {
  int out = 0, fh = 1;
  char jp = getchar();
  while ((jp > '9' || jp < '0') && jp != '-') jp = getchar();
  if (jp == '-') fh = -1, jp = getchar();
  while (jp >= '0' && jp <= '9') out = out * 10 + jp - '0', jp = getchar();
  return out * fh;
}
void print(int x) {
  if (x >= 10) print(x / 10);
  putchar('0' + x % 10);
}
void write(int x, char c) {
  if (x < 0) putchar('-'), x = -x;
  print(x);
  putchar(c);
}
const int P = 1e9 + 7;
int add(int a, int b) { return a + b >= P ? a + b - P : a + b; }
void inc(int &a, int b) { a = add(a, b); }
int mul(int a, int b) { return 1LL * a * b % P; }
int fpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
const int N = 2e5 + 10;
int fac[N], invfac[N];
int S(int n, int k) {
  if (n < k) return 0;
  int s = 0;
  for (int i = 0; i <= k; ++i) {
    int t = mul(mul(invfac[k - i], invfac[i]), fpow(k - i, n));
    if (i & 1)
      inc(s, P - t);
    else
      inc(s, t);
  }
  return s;
}
int n, k, ans = 0;
int main() {
  n = read(), k = read();
  fac[0] = 1;
  for (int i = 1; i <= n; ++i) inc(ans, read()), fac[i] = mul(fac[i - 1], i);
  invfac[n] = fpow(fac[n], P - 2);
  for (int i = n - 1; i >= 0; --i) invfac[i] = mul(invfac[i + 1], i + 1);
  int coef = add(mul(S(n - 1, k), n - 1), S(n, k));
  ans = mul(ans, coef);
  cout << ans << '\n';
  return 0;
}

```