Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.

There are n stages available. The rocket must contain exactly k of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.

For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — 26 tons.

Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.

Input

The first line of input contains two integers — n and k (1 ≤ k ≤ n ≤ 50) – the number of available stages and the number of stages to use in the rocket.

The second line contains string s, which consists of exactly n lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.

Output

Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.

Examples

Input

5 3
xyabd


Output

29

Input

7 4
problem


Output

34

Input

2 2
ab


Output

-1

Input

12 1
abaabbaaabbb


Output

1

Note

In the first example, the following rockets satisfy the condition:

  * "adx" (weight is 1+4+24=29);
  * "ady" (weight is 1+4+25=30);
  * "bdx" (weight is 2+4+24=30);
  * "bdy" (weight is 2+4+25=31).



Rocket "adx" has the minimal weight, so the answer is 29.

In the second example, target rocket is "belo". Its weight is 2+5+12+15=34.

In the third example, n=k=2, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1 << 21;
const int N = int(1e5);
const long long mod = int(1e9) + 7;
const long long inf = 1e18;
const double PI = 3.1415926536;
inline int Set(int n, int pos) { return n = n | 1 << pos; }
inline bool check(int n, int pos) { return n & 1 << pos; }
inline int Reset(int n, int pos) { return n = n & ~(1 << pos); }
inline int lastbit(int n) { return n & (-n); }
inline long long MM(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  long long c = (a * b) % mod;
  return c;
}
inline long long AM(long long a, long long b) {
  a = (a + mod) % mod;
  b = (b + mod) % mod;
  long long c = (a + b) % mod;
  return c;
}
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return (a * b) / gcd(a, b); }
inline long long power(long long a, long long b) {
  long long ans = 1;
  a = a % mod;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return ans;
}
inline long long invmod(long long a) {
  long long ans = power(a, mod - 2);
  return ans;
}
inline long long mulmod(long long a, long long b, long long c = mod) {
  long long x = 0, y = a % c;
  while (b > 0) {
    if (b % 2) x = (x + y) % c;
    y = (y * 2) % c;
    b >>= 1;
  }
  return x % c;
}
inline long long Largepower(long long a, long long b, long long c = mod) {
  long long x = 1, y = a % c;
  while (b > 0) {
    if (b & 1) x = mulmod(x, y, c);
    y = mulmod(y, y, c);
    b >>= 1;
  }
  return x;
}
inline long long Egcd(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = Egcd(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
inline int bs(int arr[], int l, int r, int item) {
  if (l == r) {
    if (arr[l] == item)
      return r;
    else
      return -1;
  }
  if (item <= arr[(l + r) / 2])
    return bs(arr, l, (l + r) / 2, item);
  else
    return bs(arr, (l + r) / 2 + 1, r, item);
}
inline long long nCrMod(long long n, long long r) {
  if (r > n / 2) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans = (ans % mod * (n - r + i + mod) % mod) % mod;
    ans = (ans % mod * invmod(i) % mod) % mod;
  }
  return ans % mod;
}
inline long long nCr(long long n, long long r) {
  if (r > n / 2) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans = (ans * (n - r + i));
    ans /= i;
  }
  return ans;
}
inline void fastin(int &x) {
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return;
}
int leap[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int nleap[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  sort(s.begin(), s.end());
  char p = s[0];
  int w = s[0] - 'a' + 1, l = 1;
  for (int i = 1; i < s.length(); i++) {
    if (l == k) break;
    if (p + 1 < s[i]) {
      w += s[i] - 'a' + 1;
      p = s[i];
      l++;
    }
  }
  if (l < k)
    cout << -1;
  else
    cout << w;
  return 0;
}
