#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
int n, k, a[N], w;
int op[N], fac[N], infac[N], inv[N];
namespace MATHEMATICS {
int add(int x, int y) {
  int ret = x + y;
  if (ret >= mod) ret -= mod;
  return ret;
}
int mi(int x, int y) {
  int ret = x - y;
  if (ret < 0) ret += mod;
  return ret;
}
void inc(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void dec(int &x, int y) {
  x -= y;
  if (x < 0) x += mod;
}
int mul(int x, int y) { return 1LL * x * y % mod; }
int F_p(int x, int y) {
  int bas = x, ret = 1;
  while (y) {
    if (y & 1) ret = mul(ret, bas);
    bas = mul(bas, bas);
    y >>= 1;
  }
  return ret;
}
}  // namespace MATHEMATICS
using namespace MATHEMATICS;
void init() {
  int i;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), inc(w, a[i]);
}
void prework() {
  int i;
  op[0] = 1;
  for (i = 1; i <= n; ++i) op[i] = mi(0, op[i - 1]);
  inv[1] = 1;
  for (i = 2; i <= n; ++i) inv[i] = mul(inv[mod % i], mod - mod / i);
  infac[0] = 1;
  for (i = 1; i <= n; ++i) infac[i] = mul(infac[i - 1], inv[i]);
  for (i = 0; i < k; ++i) fac[i] = F_p(k - i, n - 2);
}
void work() {
  int i, ans = 0;
  for (i = 0; i < k; ++i)
    inc(ans, mul(mul(mul(op[i], fac[i]), mul(infac[i], infac[k - i - 1])),
                 (k - i + n - 1)));
  printf("%d\n", mul(ans, w));
}
int main() {
  init();
  if (n == 1) {
    printf("%d\n", w);
    return 0;
  }
  prework();
  work();
  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;
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
int n, k, a[N], w;
int op[N], fac[N], infac[N], inv[N];
namespace MATHEMATICS {
int add(int x, int y) {
  int ret = x + y;
  if (ret >= mod) ret -= mod;
  return ret;
}
int mi(int x, int y) {
  int ret = x - y;
  if (ret < 0) ret += mod;
  return ret;
}
void inc(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void dec(int &x, int y) {
  x -= y;
  if (x < 0) x += mod;
}
int mul(int x, int y) { return 1LL * x * y % mod; }
int F_p(int x, int y) {
  int bas = x, ret = 1;
  while (y) {
    if (y & 1) ret = mul(ret, bas);
    bas = mul(bas, bas);
    y >>= 1;
  }
  return ret;
}
}  // namespace MATHEMATICS
using namespace MATHEMATICS;
void init() {
  int i;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), inc(w, a[i]);
}
void prework() {
  int i;
  op[0] = 1;
  for (i = 1; i <= n; ++i) op[i] = mi(0, op[i - 1]);
  inv[1] = 1;
  for (i = 2; i <= n; ++i) inv[i] = mul(inv[mod % i], mod - mod / i);
  infac[0] = 1;
  for (i = 1; i <= n; ++i) infac[i] = mul(infac[i - 1], inv[i]);
  for (i = 0; i < k; ++i) fac[i] = F_p(k - i, n - 2);
}
void work() {
  int i, ans = 0;
  for (i = 0; i < k; ++i)
    inc(ans, mul(mul(mul(op[i], fac[i]), mul(infac[i], infac[k - i - 1])),
                 (k - i + n - 1)));
  printf("%d\n", mul(ans, w));
}
int main() {
  init();
  if (n == 1) {
    printf("%d\n", w);
    return 0;
  }
  prework();
  work();
  return 0;
}

```