#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005, mod = 1000000007;
inline int read() {
  int ret, f = 1;
  char c;
  while ((c = getchar()) && (c < '0' || c > '9'))
    if (c == '-') f = -1;
  ret = c - '0';
  while ((c = getchar()) && (c >= '0' && c <= '9'))
    ret = (ret << 3) + (ret << 1) + c - '0';
  return ret * f;
}
long long add(long long x, long long y) {
  return x + y < mod ? x + y : x + y - mod;
}
long long rdc(long long x, long long y) {
  return x - y < 0 ? x - y + mod : x - y;
}
long long qpow(long long x, int y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ret;
}
int n, m;
long long ans, inv[maxn];
void init() {
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= m; ++i)
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
  for (int i = 2; i <= m; ++i) inv[i] = inv[i - 1] * inv[i] % mod;
}
int main() {
  n = read();
  m = read();
  int x;
  for (int i = 1; i <= n; ++i) {
    x = read();
    ans = add(ans, x);
  }
  if (n == 1) {
    printf("%lld", m == 1 ? ans : 0);
    return 0;
  }
  init();
  long long mul = 0, tmp;
  for (int i = 0; i < m; ++i) {
    tmp = ((i & 1) ? mod - 1 : 1);
    tmp = tmp * qpow(m - i, n - 2) % mod;
    tmp = (tmp * inv[i] % mod) * inv[m - i - 1] % mod;
    tmp = tmp * (m - i + n - 1) % mod;
    mul = add(mul, tmp);
  }
  printf("%lld", ans * mul % 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;
const int maxn = 200005, mod = 1000000007;
inline int read() {
  int ret, f = 1;
  char c;
  while ((c = getchar()) && (c < '0' || c > '9'))
    if (c == '-') f = -1;
  ret = c - '0';
  while ((c = getchar()) && (c >= '0' && c <= '9'))
    ret = (ret << 3) + (ret << 1) + c - '0';
  return ret * f;
}
long long add(long long x, long long y) {
  return x + y < mod ? x + y : x + y - mod;
}
long long rdc(long long x, long long y) {
  return x - y < 0 ? x - y + mod : x - y;
}
long long qpow(long long x, int y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ret;
}
int n, m;
long long ans, inv[maxn];
void init() {
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= m; ++i)
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
  for (int i = 2; i <= m; ++i) inv[i] = inv[i - 1] * inv[i] % mod;
}
int main() {
  n = read();
  m = read();
  int x;
  for (int i = 1; i <= n; ++i) {
    x = read();
    ans = add(ans, x);
  }
  if (n == 1) {
    printf("%lld", m == 1 ? ans : 0);
    return 0;
  }
  init();
  long long mul = 0, tmp;
  for (int i = 0; i < m; ++i) {
    tmp = ((i & 1) ? mod - 1 : 1);
    tmp = tmp * qpow(m - i, n - 2) % mod;
    tmp = (tmp * inv[i] % mod) * inv[m - i - 1] % mod;
    tmp = tmp * (m - i + n - 1) % mod;
    mul = add(mul, tmp);
  }
  printf("%lld", ans * mul % mod);
  return 0;
}

```