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 PYTHON3):

Solution:
import math

n, k = map(int, input().split())

s = list(input())

alphabet = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']

ltrs = alphabet[:k]

if k == 2:
    s1 = ("AB" * math.ceil(n/2))[:n]
    s2 = ("BA" * math.ceil(n/2))[:n]

    diff1, diff2 = 0, 0
    for i in range(n):
        if s1[i] != s[i]:
            diff1 += 1
        if s2[i] != s[i]:
            diff2 += 1

    if diff1 < diff2:
        print(diff1)
        print(s1)
    else:
        print(diff2)
        print(s2)

    exit()

c = 0
for i in range(1, n-1):
    if s[i] == s[i-1]:
        oneneed = 0
        c += 1
        for l in ltrs:
            if l != s[i-1] and l != s[i+1]:
                s[i] = l
                oneneed = 1
                break

        if oneneed == 0:
            for l in ltrs:
                if l != s[i-1]:
                    s[i] = l
                    break

if n > 1:
    if s[-1] == s[-2]:
        for l in ltrs:
            if l != s[-2]:
                s[-1] = l
                c += 1
                break

print(c)
print(''.join(s))