Problem:
Score : 300 points

Problem Statement

You are given a string S consisting of lowercase English letters.
Another string T is initially empty.
Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times:

Append one of the following at the end of T: dream, dreamer, erase and eraser.

Constraints

1≦|S|≦10^5

S consists of lowercase English letters.

Input

The input is given from Standard Input in the following format:

S

Output

If it is possible to obtain S = T, print YES. Otherwise, print NO.

Sample Input 1

erasedream

Sample Output 1

YES

Append erase and dream at the end of T in this order, to obtain S = T.

Sample Input 2

dreameraser

Sample Output 2

YES

Append dream and eraser at the end of T in this order, to obtain S = T.

Sample Input 3

dreamerer

Sample Output 3

NO

Solution in Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"bytes"
	"fmt"
	"os"
)

var sc = bufio.NewScanner(os.Stdin)

func main() {

	sc.Scan()
	S := sc.Bytes()
	if back(S) {
		fmt.Println("YES")
		return
	}

	fmt.Println("NO")
	return

	i := 0
	for i < len(S) {
		b, c := confirm(S[i:])
		if !b {
			fmt.Println("NO")
			return
		}

		i += c
	}

	fmt.Println("YES")

}

func back(b []byte) bool {
	S := reverse(b)
	divide := []([]byte){
		reverse([]byte("dream")), reverse([]byte("dreamer")), reverse([]byte("erase")), reverse([]byte("eraser")),
	}

	can := true
	for i := 0; i < len(S); {
		can2 := false
		for j := range divide {
			if bytes.HasPrefix(S[i:], divide[j]) {
				can2 = true
				i += len(divide[j])
				break
			}
		}
		if !can2 {
			return false
		}
	}
	return can
}

func reverse(b []byte) []byte {
	if len(b) == 1 {
		return b
	}
	for i, j := 0, len(b)-1; i < j; i, j = i+1, j-1 {
		b[i], b[j] = b[j], b[i]
	}

	return b
}

func try2(b []byte) bool {
	fmt.Println(string(b))
	if len(b) == 0 {
		return true
	}

	if len(b) < 5 {
		return false
	}

	if bytes.HasSuffix(b, []byte{'d', 'r', 'e', 'a', 'm'}) {
		return try2(b[:len(b)-5])
	} else if bytes.HasSuffix(b, []byte{'d', 'r', 'e', 'a', 'm', 'e', 'r'}) {
		return try2(b[:len(b)-7])
	} else if bytes.HasSuffix(b, []byte{'e', 'r', 'a', 's', 'e'}) {
		return try2(b[:len(b)-5])
	} else if bytes.HasSuffix(b, []byte{'e', 'r', 'a', 's', 'e', 'r'}) {
		return try2(b[:len(b)-6])
	}
	return false
}

func confirm(b []byte) (bool, int) {
	if has(b, []byte("dreamer")) {
		if len(b) == 7 {
			return true, 7
		}

		if b[7] == 'a' {
			return true, 5
		}

		return true, 7
	}

	if has(b, []byte("dream")) {
		return true, 5
	}

	if has(b, []byte("eraser")) {
		return true, 6
	}

	if has(b, []byte("erase")) {
		return true, 5
	}

	return false, -1
}

// b の先頭が t かどうか
func has(b []byte, t []byte) bool {
	if len(b) < len(t) {
		return false
	}

	for i := range t {
		if b[i] != t[i] {
			return false
		}
	}
	return true
}