#include <bits/stdc++.h>
using namespace std;
template <class _Tp>
_Tp gcd(_Tp a, _Tp b) {
  return (b == 0) ? (a) : (gcd(b, a % b));
}
const long long Inf = 1000000000000000000ll;
const int inf = 1000000000;
char buf[1 << 25], *p1 = buf, *p2 = buf;
inline int getc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 25, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int x = 0, f = 0;
  char c = getc();
  while (!isdigit(c)) (c == '-') && (f = 1), c = getc();
  while (isdigit(c)) x = x * 10 + (c & 15), c = getc();
  return f ? -x : x;
}
const long long mod = 1000000007;
const int N = 200005;
long long fac[N], invfac[N];
long long qpow(long long b, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * b % mod;
    b = b * b % mod;
    p >>= 1ll;
  }
  return res;
}
long long C(int n, int k) {
  return fac[n] * invfac[k] % mod * invfac[n - k] % mod;
}
long long solve(int n, int m) {
  long long res = 0;
  for (int i = 0; i < m; ++i) {
    long long t = C(m, i) * qpow(m - i, n) % mod;
    if (i & 1) t = mod - t;
    res = (res + t) % mod;
  }
  return res * invfac[m] % mod;
}
int main() {
  int n = read();
  int k = read();
  fac[0] = invfac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = (fac[i - 1] * i % mod) % mod;
    invfac[i] = qpow(fac[i], mod - 2);
  }
  long long sum = 0;
  for (int i = 1; i <= n; ++i) sum = (sum + 1ll * read() % mod) % mod;
  sum = (sum + mod) % mod;
  printf("%lld\n",
         sum * (solve(n, k) + 1ll * (n - 1) * solve(n - 1, k) % mod) % mod);
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
template <class _Tp>
_Tp gcd(_Tp a, _Tp b) {
  return (b == 0) ? (a) : (gcd(b, a % b));
}
const long long Inf = 1000000000000000000ll;
const int inf = 1000000000;
char buf[1 << 25], *p1 = buf, *p2 = buf;
inline int getc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 25, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int x = 0, f = 0;
  char c = getc();
  while (!isdigit(c)) (c == '-') && (f = 1), c = getc();
  while (isdigit(c)) x = x * 10 + (c & 15), c = getc();
  return f ? -x : x;
}
const long long mod = 1000000007;
const int N = 200005;
long long fac[N], invfac[N];
long long qpow(long long b, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * b % mod;
    b = b * b % mod;
    p >>= 1ll;
  }
  return res;
}
long long C(int n, int k) {
  return fac[n] * invfac[k] % mod * invfac[n - k] % mod;
}
long long solve(int n, int m) {
  long long res = 0;
  for (int i = 0; i < m; ++i) {
    long long t = C(m, i) * qpow(m - i, n) % mod;
    if (i & 1) t = mod - t;
    res = (res + t) % mod;
  }
  return res * invfac[m] % mod;
}
int main() {
  int n = read();
  int k = read();
  fac[0] = invfac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = (fac[i - 1] * i % mod) % mod;
    invfac[i] = qpow(fac[i], mod - 2);
  }
  long long sum = 0;
  for (int i = 1; i <= n; ++i) sum = (sum + 1ll * read() % mod) % mod;
  sum = (sum + mod) % mod;
  printf("%lld\n",
         sum * (solve(n, k) + 1ll * (n - 1) * solve(n - 1, k) % mod) % mod);
  return 0;
}

```