stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define FOR0(i,n) for(ll i=0;i<n;i++)

#define FOR1(i,n) for(ll i=1;i<=n;i++)

#define FORl(i,l,n) for(ll i=l;i<n;i++)

using namespace std;

#define pi pair<ll,ll>

#define pb push_back

#define ll long long

#define ld long double

#define ff first

#define ss second

#define mp make_pair

#define mst0(x) memset(x,0,sizeof(x));

#define vi vector<ll>

#define sync ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define endl '\n'

#define sp <<" "<<

#define intmax 1e14

#define M 1000000007



ll power(ll num,ll g,ll mod){

  if(g==0)return 1;

  if(g%2==1)return (num*power((num*num)%mod,g/2,mod))%mod;

  return power((num*num)%mod,g/2,mod);

}

 

int main(){

	ll n, k;

	cin>>n>>k;

	vector<ll> vec(n);

	FOR0(i, n)	cin>>vec[i];

	vector< vector<ll> > dp(n, vector<ll> (k+1, 0));

	vector< vector<ll> > ps(n, vector<ll> (k+1, 0));

	ps[0][0] = 1;

	FOR0(i, k+1){

		if(i <= vec[0])

			dp[0][i] = 1;

		if(i > 0)

			ps[0][i] = (ps[0][i] + dp[0][i] + ps[0][i-1]) % M;

	}

	FOR1(i, n-1){

		FOR0(j, k+1)

			dp[i][j] = (j-vec[i]-1) >= 0 ? (ps[i-1][j] - ps[i-1][j - vec[i] - 1] + M) % M : ps[i-1][j] % M;

		FOR0(j, k+1){

			ps[i][j] = j==0 ? dp[i][j] : (dp[i][j] + ps[i][j-1]) % M;

		}

	}

	cout<<dp[n-1][k]<<"\n";

}

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define FOR0(i,n) for(ll i=0;i<n;i++)

#define FOR1(i,n) for(ll i=1;i<=n;i++)

#define FORl(i,l,n) for(ll i=l;i<n;i++)

using namespace std;

#define pi pair<ll,ll>

#define pb push_back

#define ll long long

#define ld long double

#define ff first

#define ss second

#define mp make_pair

#define mst0(x) memset(x,0,sizeof(x));

#define vi vector<ll>

#define sync ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define endl '\n'

#define sp <<" "<<

#define intmax 1e14

#define M 1000000007



ll power(ll num,ll g,ll mod){

  if(g==0)return 1;

  if(g%2==1)return (num*power((num*num)%mod,g/2,mod))%mod;

  return power((num*num)%mod,g/2,mod);

}

 

int main(){

	ll n, k;

	cin>>n>>k;

	vector<ll> vec(k+1);

	vector<ll> temp(k+1);

	FOR1(i, k)	vec[i] = 0; vec[0] = 1;

	while(n--){

		ll a; cin>>a;

		ll sum = 0;

		FOR0(i, k+1){

			sum += vec[i];

			if(i-a-1 >= 0)

				sum -= vec[i-a-1];

			sum %= M;

			temp[i] = sum;

		}

		swap(vec, temp);

	}

	cout<<(vec[k] + M) % M<<"\n";

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b) for(int i = a; i < (b); ++i)

#define all(x) begin(x), end(x)

#define rall(x) rbegin(x), rend(x)

#define sz(x) (int)(x).size()

#define int long long

typedef long long ll;

typedef pair<int, int> ii;

typedef vector<int> vi;

typedef vector<bool> vb;

typedef vector<vi> vvi;



signed main() {

  const int MOD = 1e9+7;

  cin.tie(0)->sync_with_stdio(0);

  cin.exceptions(cin.failbit);

  int n, k; cin >> n >> k;

  vi v(n); for(int& i: v) cin >> i;

  vector<vi> dp(n+1, vi(k+1));

  dp[0][0] = 1;

  vi p(k+2, 1); p[0] = 0;

  vi nxt(k+2);

  rep(i,1,n+1){

    nxt.assign(k+2,0);

    rep(j,0,k+1){

      dp[i][j] = (p[j+1] - p[j-min(j, v[i-1])]) % MOD;

      nxt[j+1] = (nxt[j] + dp[i][j]) % MOD;

    }

    swap(p,nxt);

  }

  cout << (dp[n][k] + MOD) % MOD;

  return 0;

}


// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b) for(int i = a; i < (b); ++i)

#define rrep(i, a, b) for(int i = a; i >= (b); --i)

#define all(x) begin(x), end(x)

#define rall(x) rbegin(x), rend(x)

#define sz(x) (int)(x).size()

typedef long long ll;

typedef pair<int, int> ii;

typedef vector<int> vi;

typedef vector<bool> vb;

typedef vector<vi> vvi;



const int MOD = 1e9+7;



void addSelf(int &a, int b){

  a += b;

  if(a >= MOD) a -= MOD;

  if(a < 0) a += MOD;

}



int main() {

  cin.tie(0)->sync_with_stdio(0);

  cin.exceptions(cin.failbit);

  int n,k; cin >> n >> k;

  vi dp(k+1); dp[0] = 1;

  rep(i,0,n){

    int up; cin >> up;

    vi delta(k+1);

    rrep(j,k,0){

      int a = j + 1;

      int b = j + min(up, k-j);

      int tmp = dp[j];

      if(a <= b){

        addSelf(delta[a], tmp);

        if(b+1 <= k)

          addSelf(delta[b+1], -tmp);

      }

    }

    int inc = 0;

    rep(j,1,k+1){

      addSelf(inc, delta[j]);

      addSelf(dp[j],inc);

    }

  }

  cout << dp[k];

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#pragma GCC optimize(2)

#include<bits/stdc++.h>

#define LL long long

#define int LL

const LL inf=0x3f3f3f3f;

using namespace std;

template<typename tp> inline void read(tp &x)

{

	x=0; char c=getchar(); bool f=0;

	for(;c<'0'||c>'9';f|=(c=='-'),c = getchar());

	for(;c>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0',c = getchar());

	if(f) x=-x;

}

int n,m;

signed main()

{

	read(n),read(m);

	printf("%lld\n",n*m-n-m+1);

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

using namespace std;

int main()

{

	int n,m;

	scanf("%d%d",&n,&m);

	printf("%d",n*m-(n+m-1));

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

typedef long long LL;

inline LL in()

{

    LL x = 0, flag = 1; char ch = getchar();

    while (ch < '0' || ch > '9') {if (ch == '-') flag = -1; ch = getchar();}

    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();

    return x * flag;

}

LL n, m;

int main()

{

    n = in();

    m = in();

    cout << (n - 1) * (m - 1);

    return 0;

}

// Optimized solution to the same problem (faster):
#include <cstdio>



int main(){

	int a,b;

	scanf("%d%d",&a,&b);

	printf("%d",(a-1)*(b-1));

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;



int main () {

	int A, B;

	cin>>A>>B;

	cout<<(A*B)-(A+B-1)<<endl;

}




// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

    int A,B,area,lenth=1,road;



    scanf("%d %d",&A,&B);

    road=(A*lenth)+(B*lenth)-lenth*lenth;



    area=(A*B)-road;

    printf("%d\n",area);

  





}




// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

using namespace std;



int main (){

    int A,B;

    

    cin >> A >>B;



    cout << A*B-1*B-1*A+1<< endl;



    return 0;



}

// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

    int A,B,area,lenth=1,road;



    scanf("%d %d",&A,&B);

    road=(A*lenth)+(B*lenth)-lenth*lenth;



    area=(A*B)-road;

    printf("%d\n",area);

  





}




// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>



using namespace std;



int main()



{

int A,B;

scanf("%d%d",&A,&B);



int ans,rd,area;



area=A*B;

rd=A*1+B*1-1;

ans=area-rd;

printf("%d\n",ans);

printf(" \n");



}


// Optimized solution to the same problem (faster):
#include<stdio.h>

int main()

{

    int A,B,area,lenth=1,road;



    scanf("%d %d",&A,&B);

    road=(A*lenth)+(B*lenth)-lenth*lenth;



    area=(A*B)-road;

    printf("%d\n",area);

  





}




// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<string>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;



int main()

{

	string s;

	cin>>s;

	int len=s.length();

	while(len>0)

	{

		if(len>=5)

		{

			if(s.substr(len-5,5)=="dream"||s.substr(len-5,5)=="erase")

			{

				len-=5;

				continue;

			}

		}

		if(len>=6)

		{

			if(s.substr(len-6,6)=="eraser")

			{

				len-=6;

				continue;

			}

		}

		if(len>=7)

		{

			if(s.substr(len-7,7)=="dreamer")

			{

				len-=7;

				continue;

			}

		}

		if(len!=0)

		{

			cout<<"NO"<<endl;

			return 0;

		}

	}



	cout<<"YES"<<endl;

	return 0;

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>



char str[100010];

bool vis[100010] = {1};

int main() {

	scanf("%s", str);

	for (int i = 0; str[i]; ++i) {

		if (vis[i]) {

			if (strncmp(str + i, "dreamer", 7) == 0)

				vis[i + 7] = 1;

			if (strncmp(str + i, "eraser", 6) == 0)

				vis[i + 6] = 1;

			if (strncmp(str + i, "dream", 5) == 0)

				vis[i + 5] = 1;

			if (strncmp(str + i, "erase", 5) == 0)

				vis[i + 5] = 1;

		}

	}

	puts(vis[strlen(str)] ? "YES" : "NO");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h> //Ithea Myse Valgulious

namespace chtholly{

typedef long long ll;

#define re0 register int

#define rec register char

#define rel register ll

#define gc getchar

#define pc putchar

#define p32 pc(' ')

#define pl puts("")

/*By Citrus*/

inline int read(){

  int x=0,f=1;char c=gc();

  for (;!isdigit(c);c=gc()) f^=c=='-';

  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');

  return f?x:-x;

  }

inline bool read(ll &x){

  x=0;int f=1;char c=gc();

  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';

  if (!~c) return 0;

  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');

  return x=f?x:-x,1;

  }

template <typename mitsuha>

inline int write(mitsuha x){

  if (!x) return pc(48);

  if (x<0) x=-x,pc('-');

  int bit[20],i,p=0;

  for (;x;x/=10) bit[++p]=x%10;

  for (i=p;i;--i) pc(bit[i]+48);

  }

inline char fuhao(){

  char c=gc();

  for (;isspace(c);c=gc());

  return c;

  }

}using namespace chtholly;

using namespace std;

string s;

int main(){

ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);

cin>>s;

for (;!s.empty();){

  string tmp=s.substr(s.size()-5,5);

  if (tmp=="erase"||tmp=="dream"){

    s.erase(s.size()-5,5);

    continue;

    }

  tmp=s.substr(s.size()-6,6);

  if (tmp=="eraser"){

    s.erase(s.size()-6,6);

    continue;

    }

  tmp=s.substr(s.size()-7,7);

  if (tmp=="dreamer"){

    s.erase(s.size()-7,7);

    continue;

    }

  return puts("NO"),0;

  }

puts("YES");

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>



char str[100010];

bool vis[100010] = {1};

int main() {

	scanf("%s", str);

	for (int i = 0; str[i]; ++i) {

		if (vis[i]) {

			if (strncmp(str + i, "dreamer", 7) == 0)

				vis[i + 7] = 1;

			if (strncmp(str + i, "eraser", 6) == 0)

				vis[i + 6] = 1;

			if (strncmp(str + i, "dream", 5) == 0)

				vis[i + 5] = 1;

			if (strncmp(str + i, "erase", 5) == 0)

				vis[i + 5] = 1;

		}

	}

	puts(vis[strlen(str)] ? "YES" : "NO");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

#include <iomanip>

#define ALL(x) (x).begin(), (x).end()

#define rt return

#define dll(x) scanf("%I64d",&x)

#define xll(x) printf("%I64d\n",x)

#define sz(a) int(a.size())

#define all(a) a.begin(), a.end()

#define rep(i,x,n) for(int i=x;i<n;i++)

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define eps 1e-6

#define gg(x) getInt(&x)

#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;

using namespace std;

typedef long long ll;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}

inline void getInt(int* p);

const int maxn=1000010;

const int inf=0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/



int main()

{

    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);

	//freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);

	string s;

	cin>>s;

	while(1)

	{

		int len=s.length();

		if(len<5)

		{

			cout<<"NO"<<endl;

			break;

		}



        if(len>=11&&s.substr(0,11)=="dreameraser")

		{

			s.erase(0,11);

		}

		else if(len>=10&&s.substr(0,10)=="dreamerase")

		{

			s.erase(0,10);

		}else if(s.substr(0,7)=="dreamer")

		{

			s.erase(0,7);

		}else if(s.substr(0,5)=="dream")

		{

			s.erase(0,5);

		}else if(s.substr(0,6)=="eraser")

		{

			s.erase(0,6);

		}else if(s.substr(0,5)=="erase")

		{

			s.erase(0,5);

		}



		if(s.length()==0)

		{

			cout<<"YES"<<endl;

			break;

		}

		if(len==s.length())

        {

            cout<<"NO"<<endl;

			break;

        }

	}

    return 0;

}



inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 - ch + '0';

        }

    }

    else {

        *p = ch - '0';

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 + ch - '0';

        }

    }

}




// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>



char str[100010];

bool vis[100010] = {1};

int main() {

	scanf("%s", str);

	for (int i = 0; str[i]; ++i) {

		if (vis[i]) {

			if (strncmp(str + i, "dreamer", 7) == 0)

				vis[i + 7] = 1;

			if (strncmp(str + i, "eraser", 6) == 0)

				vis[i + 6] = 1;

			if (strncmp(str + i, "dream", 5) == 0)

				vis[i + 5] = 1;

			if (strncmp(str + i, "erase", 5) == 0)

				vis[i + 5] = 1;

		}

	}

	puts(vis[strlen(str)] ? "YES" : "NO");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <cstring>

#include <algorithm>



char s[4][200] = {"dreamer", "dream", "erase", "eraser"};

char str[200000];

int l[4] = {7, 5, 5, 6};

int len;

bool flg;



void dfs(int idx) {

//    std::cout << cur << " " <<ll <<  std::endl;

    if (flg) return;

    if (idx == len) {

        flg = true;

        return;

    } else if (idx > len) return;

    for (int j = 0; j < 4; ++j) {

        bool fl = true;

        for (int i = 0; i < l[j]; ++i) {

            if (s[j][i] != str[idx + i]) {

                fl = false;

                break;

            }

        }

        if (fl) dfs(idx + l[j]);

    }

}



int main() {

    std::cin >> str;

    len = strlen(str);

    dfs(0);

    puts(flg ? "YES" : "NO");

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>



char str[100010];

bool vis[100010] = {1};

int main() {

	scanf("%s", str);

	for (int i = 0; str[i]; ++i) {

		if (vis[i]) {

			if (strncmp(str + i, "dreamer", 7) == 0)

				vis[i + 7] = 1;

			if (strncmp(str + i, "eraser", 6) == 0)

				vis[i + 6] = 1;

			if (strncmp(str + i, "dream", 5) == 0)

				vis[i + 5] = 1;

			if (strncmp(str + i, "erase", 5) == 0)

				vis[i + 5] = 1;

		}

	}

	puts(vis[strlen(str)] ? "YES" : "NO");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<iostream>

#include<string>

#include<cmath>

#include<algorithm>

#include<queue>

#include<set>

#include<stack>

#include<cstring>

#include<map>

#include<vector>

#define ll long long int

using namespace std;

int main()

{

   string s;

   cin>>s;

   ll t;

    while((t=s.find("eraser"))!=s.npos)

   {

       s=s.replace(t,6,"");

   }

     while((t=s.find("erase"))!=s.npos)

   {

       s=s.replace(t,5,"");

   }

    while((t=s.find("dreamer"))!=s.npos)

   {

       s=s.replace(t,7,"");

   }

   while((t=s.find("dream"))!=s.npos)

   {

       s=s.replace(t,5,"");

   }

   if(s.length()==0)

    cout<<"YES";

   else

    cout<<"NO";

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>



char str[100010];

bool vis[100010] = {1};

int main() {

	scanf("%s", str);

	for (int i = 0; str[i]; ++i) {

		if (vis[i]) {

			if (strncmp(str + i, "dreamer", 7) == 0)

				vis[i + 7] = 1;

			if (strncmp(str + i, "eraser", 6) == 0)

				vis[i + 6] = 1;

			if (strncmp(str + i, "dream", 5) == 0)

				vis[i + 5] = 1;

			if (strncmp(str + i, "erase", 5) == 0)

				vis[i + 5] = 1;

		}

	}

	puts(vis[strlen(str)] ? "YES" : "NO");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <algorithm>

#include <iostream>

using namespace std;



const int MAX_N = 1e5 + 5, MOD = 1e9 + 7;

int n;

bool dp[MAX_N];

string s;



int main() {

    cin >> s;

   	for (int i = 1; i <= s.size(); i++) {

   		if (i < 5) {

   			dp[i] = false;

   			continue;

   		}



   		if (i == 5) {

   			if (s[0] == 'd' && s[1] == 'r' && s[2] == 'e' && s[3] == 'a' && s[4] == 'm')

   				dp[i] = true;

   			if (s[0] == 'e' && s[1] == 'r' && s[2] == 'a' && s[3] == 's' && s[4] == 'e')

   				dp[i] = true;

   			continue;

   		}



   		if (i == 6) {

   			if (s[0] == 'e' && s[1] == 'r' && s[2] == 'a' && s[3] == 's' && s[4] == 'e' && s[5] == 'r')

   				dp[i] = true;

   			continue;

   		}



   		if (i == 7) {

   			if (s[0] == 'd' && s[1] == 'r' && s[2] == 'e' && s[3] == 'a' && s[4] == 'm' && s[5] == 'e' && s[6] == 'r')

   				dp[i] = true;

   			continue;

   		}



   		if (s[i - 1] == 'm') {

   			if (s[i - 5] == 'd' && s[i - 4] == 'r' && s[i - 3] == 'e' && s[i - 2] == 'a' && s[i - 1] == 'm')

   				dp[i] = dp[i - 5];



   			continue;

   		}





   		if (s[i - 1] == 'e') {

   			if (s[i - 5] == 'e' && s[i - 4] == 'r' && s[i - 3] == 'a' && s[i - 2] == 's' && s[i - 1] == 'e')

   				dp[i] = dp[i - 5];



   			continue;

   		}



   		if (s[i - 1] != 'r')

   			continue;



   		if (s[i - 3] == 'm') {

   			if (s[i - 7] == 'd' && s[i - 6] == 'r' && s[i - 5] == 'e' && s[i - 4] == 'a' && s[i - 3] == 'm' && s[i - 2] == 'e' && s[i - 1] == 'r')

   				dp[i] = dp[i - 7];



   			continue;

   		}

   		if (s[i - 3] == 's') {

   			if (s[i - 6] == 'e' && s[i - 5] == 'r' && s[i - 4] == 'a' && s[i - 3] == 's' && s[i - 2] == 'e' && s[i - 1] == 'r')

   				dp[i] = dp[i - 6];



   			continue;

   		}

   	}



   	cout << (dp[s.size()]? "YES": "NO");

    return 0;

}

// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>



char str[100010];

bool vis[100010] = {1};

int main() {

	scanf("%s", str);

	for (int i = 0; str[i]; ++i) {

		if (vis[i]) {

			if (strncmp(str + i, "dreamer", 7) == 0)

				vis[i + 7] = 1;

			if (strncmp(str + i, "eraser", 6) == 0)

				vis[i + 6] = 1;

			if (strncmp(str + i, "dream", 5) == 0)

				vis[i + 5] = 1;

			if (strncmp(str + i, "erase", 5) == 0)

				vis[i + 5] = 1;

		}

	}

	puts(vis[strlen(str)] ? "YES" : "NO");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define ll long long



using namespace std;



string s[5]={"dream","dreamer","erase","eraser"},a;



void dfs(int pos)

{

	if(pos==a.size())

	{

		printf("YES");

		exit(0);

	}

	if(a.substr(pos,5)==s[0])dfs(pos+5);

	if(a.substr(pos,5)==s[2])dfs(pos+5);

	if(a.substr(pos,6)==s[3])dfs(pos+6);

	if(a.substr(pos,7)==s[1])dfs(pos+7);

}



int main()

{

	cin>>a;

	dfs(0);

	printf("NO");

	return 0;

}

// Optimized solution to the same problem (faster):
#include<cstdio>

#include<string.h>

using namespace std;

char s[1000000];

int len;

bool b=true;

int main(){

	scanf("%s",&s);

	len=strlen(s);

	for (int i=0;i<len;i++){

		if ((s[i]=='d')&&(s[i+1]=='r')&&(s[i+2]=='e')&&(s[i+3]=='a')&&(s[i+4]=='m')&&(s[i+5]=='e')&&(s[i+6]=='r')){

			i+=6;

			if ((s[i-1]=='e')&&(s[i]=='r')&&(s[i+1]=='a')&&(s[i+2]=='s')&&(s[i+3]=='e')&&(s[i+4]=='r')){

				i+=4;

			}else if ((s[i-1]=='e')&&(s[i]=='r')&&(s[i+1]=='a')&&(s[i+2]=='s')&&(s[i+3]=='e')){

				i+=3;

			}

		}else if ((s[i]=='d')&&(s[i+1]=='r')&&(s[i+2]=='e')&&(s[i+3]=='a')&&(s[i+4]=='m')){

			i+=4;

		}else if ((s[i]=='e')&&(s[i+1]=='r')&&(s[i+2]=='a')&&(s[i+3]=='s')&&(s[i+4]=='e')&&(s[i+5]=='r')){

			i+=5;

		}else if ((s[i]=='e')&&(s[i+1]=='r')&&(s[i+2]=='a')&&(s[i+3]=='s')&&(s[i+4]=='e')){

			i+=4;

		}else {

			b=false;

		}

		if (b==false){

			break;

		}

	}

if (b==true){

	printf("YES");

}else{

	printf("NO");

}

return 0;}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>



using namespace std;



int main()

{

    int i =0;

    string s;

    cin >> s;

    bool flag = true;

    while(i<s.length())

    {

        if(s[i]=='e')

        {

            if(i+5<s.length()&&s[i+1]=='r'&&s[i+2]=='a'&&s[i+3]=='s'&&s[i+4]=='e'&&s[i+5]=='r')

            {

                i+=6;

            }

            else if(i+4<s.length()&&s[i+1]=='r'&&s[i+2]=='a'&&s[i+3]=='s'&&s[i+4]=='e')

            {

                i+=5;

            }

            else if(i+1<s.length()&&s[i+1]=='r'&&i>1&&s[i-1]=='m')

            {

                i+=2;

            }

            else{

                flag = false;

                break;

            }

        }

        else if(s[i]=='d')

        {



            if(i+4<s.length()&&s[i+1]=='r'&&s[i+2]=='e'&&s[i+3]=='a'&&s[i+4]=='m')

            {

                i+=5;

            }

            else{

                flag = false;

                break;

            }

        }

        else{

            flag = false;

            break;

        }

    }

    flag?cout<<"YES"<<endl:cout<<"NO"<<endl;

    return 0;

}






// Optimized solution to the same problem (faster):
#include<cstdio>

#include<string.h>

using namespace std;

char s[1000000];

int len;

bool b=true;

int main(){

	scanf("%s",&s);

	len=strlen(s);

	for (int i=0;i<len;i++){

		if ((s[i]=='d')&&(s[i+1]=='r')&&(s[i+2]=='e')&&(s[i+3]=='a')&&(s[i+4]=='m')&&(s[i+5]=='e')&&(s[i+6]=='r')){

			i+=6;

			if ((s[i-1]=='e')&&(s[i]=='r')&&(s[i+1]=='a')&&(s[i+2]=='s')&&(s[i+3]=='e')&&(s[i+4]=='r')){

				i+=4;

			}else if ((s[i-1]=='e')&&(s[i]=='r')&&(s[i+1]=='a')&&(s[i+2]=='s')&&(s[i+3]=='e')){

				i+=3;

			}

		}else if ((s[i]=='d')&&(s[i+1]=='r')&&(s[i+2]=='e')&&(s[i+3]=='a')&&(s[i+4]=='m')){

			i+=4;

		}else if ((s[i]=='e')&&(s[i+1]=='r')&&(s[i+2]=='a')&&(s[i+3]=='s')&&(s[i+4]=='e')&&(s[i+5]=='r')){

			i+=5;

		}else if ((s[i]=='e')&&(s[i+1]=='r')&&(s[i+2]=='a')&&(s[i+3]=='s')&&(s[i+4]=='e')){

			i+=4;

		}else {

			b=false;

		}

		if (b==false){

			break;

		}

	}

if (b==true){

	printf("YES");

}else{

	printf("NO");

}

return 0;}

// Fastest known solution for this problem: