Problem:
Score : 300 points

Problem Statement

We will call a string obtained by arranging the characters contained in a string a in some order, an anagram of a.

For example, greenbin is an anagram of beginner. As seen here, when the same character occurs multiple times, that character must be used that number of times.

Given are N strings s_1, s_2, \ldots, s_N. Each of these strings has a length of 10 and consists of lowercase English characters. Additionally, all of these strings are distinct. Find the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Constraints

2 \leq N \leq 10^5

s_i is a string of length 10.

Each character in s_i is a lowercase English letter.

s_1, s_2, \ldots, s_N are all distinct.

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N

Output

Print the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Sample Input 1

3
acornistnt
peanutbomb
constraint

Sample Output 1

1

s_1 = acornistnt is an anagram of s_3 = constraint. There are no other pairs i, j such that s_i is an anagram of s_j, so the answer is 1.

Sample Input 2

2
oneplustwo
ninemodsix

Sample Output 2

0

If there is no pair i, j such that s_i is an anagram of s_j, print 0.

Sample Input 3

5
abaaaaaaaa
oneplustwo
aaaaaaaaba
twoplusone
aaaabaaaaa

Sample Output 3

4

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

import com.sun.org.apache.xpath.internal.operations.Bool
import java.lang.Math.*

const val MOD = 1000000007L

fun main(args: Array<String>) {
    val n = readInt()

    val s = mutableListOf<String>()
    n.rep {
        val _s = readLine()!!
        val ns = _s.map { it - '0' }.sorted()
        val ns2 = ns.joinToString(separator = "")
        s.add(ns2)
    }

    s.sort()

    var prev = ""
    var continueNum = 0
    var ans = 0
    s.forEach {
        if (it == prev) {
            ans++
            ans += continueNum
            continueNum++
        } else {
            continueNum = 0
        }
        prev = it
    }


    println(ans)
}

fun readList() = readLine()!!.split(' ')
fun readInt() = readLine()!!.split(' ').map(String::toInt).first()
fun readIntList() = readList().map(String::toInt)
fun readLongList() = readList().map(String::toLong)
fun readDoubleList() = readList().map(String::toDouble)

fun readPair(): Pair<String, String> {
    val splite = readLine()!!.split(' ')
    return Pair(splite[0], splite[1])
}

fun <T> List<T>.copy() = this.map { it }

inline fun <reified T> dualArray(h: Int, w: Int, init: (Int, Int) -> T) {
    Array(h) { i ->
        return@Array Array(w) { j ->
            return@Array init(i, j)
        }
    }
}

// 速度のためforEachは避ける
inline fun Int.rep(start: Int = 0, shiftLast: Boolean = false, body: (Int) -> Unit) {
    for (i in start until if (shiftLast) this + start else this) {
        body(i)
    }
}

inline fun Long.rep(start: Long = 0, shiftLast: Boolean = false, body: (Long) -> Unit) {
    for (i in start until if (shiftLast) this + start else this) {
        body(i)
    }
}

// 約数のList
fun divisor(value: Long): List<Long> {
    val max = sqrt(value.toDouble()).toLong()
    val former = mutableListOf<Long>()
    val latter = mutableListOf<Long>()
    (1..max).filter { value % it == 0L }
        .forEach {
            former.add(it)
            latter.add(value / it)
        }
    former.addAll(latter.reversed())
    return former
}

// 範囲内の素数を取得
// fromだけ指定すると戻り値の個数で素数判定ができる
fun prime(from: Long, to: Long = from): List<Long> {
    return (from..to).filter { i ->
        val max = Math.sqrt(i.toDouble()).toLong()
        (2..max).all { j -> i % j != 0L }
    }
}

// 素因数分解
fun decom(value: Long): List<Long> {
    if (value == 1L) return listOf(1)
    val max = Math.sqrt(value.toDouble()).toLong()
    return prime(2, max).filter { value % it == 0L }
}

// 最大公約数
fun gcd(a: Long, b: Long): Long {
    return if (a % b == 0L) b else gcd(b, a % b)
}

// 文字列を入れ替え
fun swap(base: String, a: String, b: String): String {
    return base.map {
        when (it) {
            a.toCharArray()[0] -> b
            b.toCharArray()[0] -> a
            else -> it.toString()
        }
    }.joinToString()
}

// 二分探索
fun List<Long>.binarySearch(target: Long, _start: Int = 0, _end: Int = this.size): Int {
    var start = (_start - 1).toLong()
    var end = _end.toLong()
    while (end - start > 1) {
        val mid = (start + end) / 2
        if (this[mid.toInt()] < target) {
            start = mid
        } else {
            end = mid
        }
    }
    return end.toInt()
}

// bit全探索
inline fun Int.bitAllSearch(body: (Array<Int>) -> Unit) {
    for (bit in 0 until (1 shl this)) {
        val s = mutableListOf<Int>()
        for (i in 0 until this) {
            if ((bit and (1 shl i)) shr i == 1) { // i が bit に入るかどうか
                s.add(i)
            }
        }
        body(s.toTypedArray())
    }
}