#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long w = 0, h = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') h = -h;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    w = w * 10 + ch - '0';
    ch = getchar();
  }
  return w * h;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long mod = 1000000007;
long long n, k, a[2000010], ans, sum;
long long val[2000010], cnt[2000010], s[2000010];
map<long long*, bool> mp;
long long qpow(long long b, long long k) {
  long long s = 1;
  while (k) {
    if (k & 1) s = s * b % mod;
    b = b * b % mod;
    k >>= 1;
  }
  return s % mod;
}
long long inv(long long a) { return qpow(a, mod - 2); }
long long C(long long n, long long m) {
  return cnt[n] * inv(cnt[m]) % mod * inv(cnt[n - m]) % mod;
}
signed main() {
  n = read();
  k = read();
  for (long long i = 1; i <= n; i++) a[i] = read(), sum = (sum + a[i]) % mod;
  cnt[0] = 1;
  for (long long i = 1; i <= n; i++) cnt[i] = cnt[i - 1] * i % mod;
  long long ans1 = 0, ans2 = 0;
  for (long long i = 0; i <= k; i++)
    ans1 =
        (ans1 + qpow(-1, i) * C(k, i) % mod * qpow(k - i, n) % mod + mod) % mod;
  ans1 = ans1 * inv(cnt[k]) % mod;
  for (long long i = 0; i <= k; i++)
    ans2 =
        (ans2 + qpow(-1, i) * C(k, i) % mod * qpow(k - i, n - 1) % mod + mod) %
        mod;
  ans2 = ans2 * inv(cnt[k]) % mod;
  printf("%lld\n", sum * (ans1 + ans2 * (n - 1) % mod) % mod);
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
inline long long read() {
  long long w = 0, h = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') h = -h;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    w = w * 10 + ch - '0';
    ch = getchar();
  }
  return w * h;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long mod = 1000000007;
long long n, k, a[2000010], ans, sum;
long long val[2000010], cnt[2000010], s[2000010];
map<long long*, bool> mp;
long long qpow(long long b, long long k) {
  long long s = 1;
  while (k) {
    if (k & 1) s = s * b % mod;
    b = b * b % mod;
    k >>= 1;
  }
  return s % mod;
}
long long inv(long long a) { return qpow(a, mod - 2); }
long long C(long long n, long long m) {
  return cnt[n] * inv(cnt[m]) % mod * inv(cnt[n - m]) % mod;
}
signed main() {
  n = read();
  k = read();
  for (long long i = 1; i <= n; i++) a[i] = read(), sum = (sum + a[i]) % mod;
  cnt[0] = 1;
  for (long long i = 1; i <= n; i++) cnt[i] = cnt[i - 1] * i % mod;
  long long ans1 = 0, ans2 = 0;
  for (long long i = 0; i <= k; i++)
    ans1 =
        (ans1 + qpow(-1, i) * C(k, i) % mod * qpow(k - i, n) % mod + mod) % mod;
  ans1 = ans1 * inv(cnt[k]) % mod;
  for (long long i = 0; i <= k; i++)
    ans2 =
        (ans2 + qpow(-1, i) * C(k, i) % mod * qpow(k - i, n - 1) % mod + mod) %
        mod;
  ans2 = ans2 * inv(cnt[k]) % mod;
  printf("%lld\n", sum * (ans1 + ans2 * (n - 1) % mod) % mod);
  return 0;
}

```