Problem: A colored stripe is represented by a horizontal row of n square cells, each cell is pained one of k colors. Your task is to repaint the minimum number of cells so that no two neighbouring cells are of the same color. You can use any color from 1 to k to repaint the cells.

Input

The first input line contains two integers n and k (1 ≤ n ≤ 5·105; 2 ≤ k ≤ 26). The second line contains n uppercase English letters. Letter "A" stands for the first color, letter "B" stands for the second color and so on. The first k English letters may be used. Each letter represents the color of the corresponding cell of the stripe.

Output

Print a single integer — the required minimum number of repaintings. In the second line print any possible variant of the repainted stripe.

Examples

Input

6 3
ABBACC


Output

2
ABCACA


Input

3 2
BBB


Output

1
BAB

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T, class first>
inline bool getbit(T a, first i) {
  T t = 1;
  return ((a & (t << i)) > 0);
}
template <class T, class first>
inline T setbit(T a, first i) {
  T t = 1;
  return (a | (t << i));
}
template <class T, class first>
inline T resetbit(T a, first i) {
  T t = 1;
  return (a & (~(t << i)));
}
template <class T, class first>
inline T togglebit(T a, first i) {
  T t = 1;
  return (a ^ (t << i));
}
template <typename T>
T POW(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * POW(B, P - 1);
  else
    return (POW(B, P / 2) * POW(B, P / 2));
}
template <typename T>
T BIGMOD(T b, T p, T m) {
  if (p == 0)
    return 1;
  else if (!(p & 1))
    return (Bigmod(b, p / 2, m) * Bigmod(b, p / 2, m)) % m;
  else
    return ((b % m) * Bigmod(b, p - 1, m)) % m;
}
template <typename T>
T Bigmod(T base, T power, T mod) {
  T ret = 1;
  while (power) {
    if (power & 1) ret = (ret * base) % mod;
    base = (base * base) % mod;
    power >>= 1;
  }
  return ret;
}
template <typename T>
T ModInverse(T number, T mod) {
  return Bigmod(number, mod - 2, mod);
}
template <typename T>
T GCD(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
T LCM(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <typename T>
T EUCLIDE(T a, T b, T &x, T &y) {
  if (a < 0) {
    T d = euclide(-a, b, x, y);
    x = -x;
    return d;
  }
  if (b < 0) {
    T d = euclide(a, -b, x, y);
    y = -y;
    return d;
  }
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    T d = euclide(b, a % b, x, y);
    T t = x;
    x = y;
    y = t - (a / b) * y;
    return d;
  }
}
template <typename T>
T ABS(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
double DEG(double x) { return (180.0 * x) / ((2.0 * acos(0.0))); }
double RAD(double x) { return (x * (double)(2.0 * acos(0.0))) / (180.0); }
template <typename T>
T DIS(T x1, T y1, T x2, T y2) {
  return sqrt((double)((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)));
}
template <typename T>
T ANGLE(T x1, T y1, T x2, T y2) {
  return atan(double(y1 - y2) / double(x1 - x2));
}
template <typename T>
long long isLeft(T a, T b, T c) {
  return (c.x - a.x) * (b.y - a.y) - (b.x - a.x) * (c.y - a.y);
}
void P_ARR(int *ar, int a, int b) {
  if (a > b) swap(a, b);
  if (a <= b) cout << ar[a];
  for (int i = a + 1; i <= b; i++) cout << " " << ar[i];
  cout << endl;
}
int dp[500007][27];
int path[500007][27];
char s[500007];
int l;
int n;
char res[500007];
int call(int pos, int pre) {
  if (pos == n) return 0;
  int &ret = dp[pos][pre];
  int &p = path[pos][pre];
  if (ret != -1) return ret;
  ret = 1000000000;
  int x = s[pos] - 'A';
  if (pre == x) {
    for (int i = 0; i <= l - 1; i++)
      if (i != x) {
        int rt = call(pos + 1, i) + 1;
        if (rt < ret) ret = rt, p = i, res[pos] = i + 'A';
      }
  } else
    ret = call(pos + 1, x), p = x, res[pos] = x + 'A';
  return ret;
}
void solve(int pos, int pre) {
  if (pos) printf("%c", pre + 'A');
  if (pos == n) return;
  solve(pos + 1, path[pos][pre]);
}
int main() {
  scanf("%d%d", &(n), &(l));
  scanf("%s", s);
  memset(dp, -1, sizeof dp);
  int suru = 26;
  int r1 = call(0, 26);
  int r2 = call(0, s[0] - 'A');
  cout << min(r1, r2) << endl;
  if (r1 < r2)
    solve(0, 26);
  else
    solve(0, s[0] - 'A');
  return 0;
}