Bees Alice and Alesya gave beekeeper Polina famous card game "Set" as a Christmas present. The deck consists of cards that vary in four features across three options for each kind of feature: number of shapes, shape, shading, and color. In this game, some combinations of three cards are said to make up a set. For every feature — color, number, shape, and shading — the three cards must display that feature as either all the same, or pairwise different. The picture below shows how sets look.

<image>

Polina came up with a new game called "Hyperset". In her game, there are n cards with k features, each feature has three possible values: "S", "E", or "T". The original "Set" game can be viewed as "Hyperset" with k = 4.

Similarly to the original game, three cards form a set, if all features are the same for all cards or are pairwise different. The goal of the game is to compute the number of ways to choose three cards that form a set.

Unfortunately, winter holidays have come to an end, and it's time for Polina to go to school. Help Polina find the number of sets among the cards lying on the table.

Input

The first line of each test contains two integers n and k (1 ≤ n ≤ 1500, 1 ≤ k ≤ 30) — number of cards and number of features.

Each of the following n lines contains a card description: a string consisting of k letters "S", "E", "T". The i-th character of this string decribes the i-th feature of that card. All cards are distinct.

Output

Output a single integer — the number of ways to choose three cards that form a set.

Examples

Input


3 3
SET
ETS
TSE


Output


1

Input


3 4
SETE
ETSE
TSES


Output


0

Input


5 4
SETT
TEST
EEET
ESTE
STES


Output


2

Note

In the third example test, these two triples of cards are sets:

  1. "SETT", "TEST", "EEET" 
  2. "TEST", "ESTE", "STES" 
import sys 
# sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
"""*******************************************************"""
def main():
    n,m=cin()
    a=[]
    d={}
    for i in range(n):
        s=sin()
        a.append(s)
        if s not in d:
            d[s]=1
        else:
            d[s]+=1
    ans=0
    e={}
    for i in range(n):
        for j in range(i+1,n):
            x=""
            for k in range(m):
                if(a[i][k]==a[j][k]):
                    x+=a[i][k]
                else:
                    if(a[i][k]=="S"):
                        if(a[j][k]=="E"):
                            x+="T"
                        else:
                            x+="E"
                    elif(a[i][k]=="E"):
                        if(a[j][k]=="S"):
                            x+="T"
                        else:
                            x+="S"
                    else:
                        if(a[j][k]=="S"):
                            x+="E"
                        else:
                            x+="S"
            if (x in d and d[x]>0):
                # print(x,d,i,j)
                p=""
                aa=[]
                aa.append(x)
                aa.append(a[i])
                aa.append(a[j])
                aa.sort()
                p+=aa[0]+"-"+aa[1]+"-"+aa[2]
                # print(p)
                if p not in e:
                    ans+=1
                    e[p]=1
    print(ans)
                    
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()