// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



namespace SCC {

	const int N = 110000;

	vector<int> con[N], rcon[N];

	int n, vis[N], visT;

	int id[N], idn;

	void init(int nn = 0) {

		n = nn, visT = idn = 0;

		for (int i = 1; i <= n; i++) con[i].clear(), rcon[i].clear();

	}

	void add_edge(int u, int v) {

		con[u].push_back(v);

		rcon[v].push_back(u);

	}

	void dfs(int u) {

		vis[u] = 1;

		for (int v : con[u]) if (!vis[v]) dfs(v);

		id[idn++] = u;

	}

	void rdfs(int u) {

		vis[u] = visT;

		for (int v : rcon[u]) if (!vis[v]) rdfs(v);

	}

	void SCC() {

		for (int i = 1; i <= n; i++) vis[i] = 0;

		for (int i = 1; i <= n; i++) if (!vis[i]) dfs(i);

		for (int i = 1; i <= n; i++) vis[i] = 0;

		for (int i = idn - 1; i >= 0; i--) if (!vis[id[i]]) visT++, rdfs(id[i]);

	}

};



const int N = 1100;

const int M = 220000;



vector<int> con[N];



struct Edge {

	int u, v;

	Edge(int u = 0, int v = 0) : u(u), v(v) {}

} E[M];



int adj[N][N];

int vis[N], visT;

int son[2][N][N];



inline bool Find(int u, int tar) {

	if (adj[u][tar] > 1) return true;

	return son[0][u][tar] != tar || son[1][u][tar] != tar;

}



void dfs(int u, int id, int rt, int __rt) {

	vis[u] = visT;

	son[id][__rt][u] = rt;

	for (int v : con[u]) if (vis[v] != visT) dfs(v, id, rt, __rt);

}



void init(int n) {

	for (int i = 1; i <= n; i++) {

		visT++;

		vis[i] = visT;

		for (int v : con[i]) if (vis[v] != visT) dfs(v, 0, v, i);

		reverse(con[i].begin(), con[i].end());

		visT++;

		vis[i] = visT;

		for (int v : con[i]) if (vis[v] != visT) dfs(v, 1, v, i);

	}

}



int main() {

	

	ios::sync_with_stdio(0);

	cin.tie(0), cout.tie(0);

	

	int n, m; cin >> n >> m;

	SCC::init(n + 10);

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

		int u, v; cin >> u >> v;

		E[i] = Edge(u, v);

		con[u].push_back(v);

		adj[u][v]++;

		SCC::add_edge(u, v);

	}

	SCC::SCC();

	

	init(n);

	

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

		auto a = SCC::vis[E[i].u], b = SCC::vis[E[i].v];

		if (a == b) {

			if (Find(E[i].u, E[i].v)) cout << "same\n";

			else cout << "diff\n";

		} else {

			if (Find(E[i].u, E[i].v)) cout << "diff\n";

			else cout << "same\n";

		}

	}

	

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 8.943889108724344x):
#include <bits/stdc++.h>

using namespace std;

#define N 1010

#define M 200010

vector <int> G[N];

int to[M],tag[N],q[2*N];

bool ans1[M],ans2[M];

int read(){

	int x=0,f=1;

	char ch=getchar();

	while (ch<'0' || ch>'9'){

		if (ch=='-') f=-1;

		ch=getchar();

	}

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

		x=x*10+ch-'0';ch=getchar();

	}

	return x*f;

}

signed main(){

	int n=read(),m=read();

	for (int i=1;i<=m;++i){

		int x=read();to[i]=read();

		G[x].push_back(i);

	}

	for (int i=1;i<=n;++i){

		for (int j=1;j<=n;++j) tag[j]=0;

		int t=1,w=0;

		for (auto &x:G[i]){

			q[++w]=to[x];tag[to[x]]=x;

		}

		while (t<=w){

			int x=q[t++],y=tag[x];

			for (auto &p:G[x]){

				int tmp=to[p];

				if (tmp!=i){

					if (tag[tmp]!=-1){

						if (tag[tmp]==0){

							q[++w]=tmp;tag[tmp]=y;

						}else if (tag[tmp]!=y){

							q[++w]=tmp;tag[tmp]=-1;

						}

					}

				}else ans1[p]=1;

			}

		}

		for (auto &x:G[i]){

			if (tag[to[x]]==-1) ans2[x]=1;

		}

	}

	for (int i=1;i<=m;++i) puts(ans1[i]^ans2[i]?"diff":"same");

	return 0; 

}