//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%lld", &(n))
#define nai(a, n) for (int i = 0; i < (n); i++) ni(a[i])
#define nal(a, n) for (int i = 0; i < (n); i++) nl(a[i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%lld\n", (n))
#define pii pair<int, int>
#define pil pair<int, long long>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vil vector<pil>
#define vll vector<pll>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef cc_hash_table<int,int,hash<int>> ht;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> oset;
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 2e5 + 5;
const double eps = 1e-9;
int a[MAXN], vis[MAXN], cnt = 5;
char ans[MAXN];
vii adj[MAXN];
vi comp;
int n, m;
bool dfs(int u, int p) {
if (u == p)
return 1;
if (vis[u % n] == cnt)
return 0;
vis[u % n] = cnt;
for (pii nx: adj[u]) {
comp.pb(nx.se);
if (dfs(nx.fi, p))
return 1;
comp.pop_back();
}
return 0;
}
bool ok(int cur) {
vis[cur] = ++cnt;
for (pii nx: adj[cur]) {
if (nx.fi == cur-1) continue;
comp.pb(nx.se);
if (dfs(nx.fi, cur + n))
return 1;
comp.pop_back();
}
return 0;
}
int main() {
scanf("%d %d", &n, &m);
for (int i = 0; i < m; i++) {
int l, r;
scanf("%d %d", &l, &r);
if (l > r) {
a[l]++, a[n + 1]--;
a[1]++, a[r + 1]--;
l--, r += n;
adj[l].pb({r, i});
adj[l+n].pb({2*n, i});
} else {
a[l]++, a[r + 1]--;
l--;
adj[l].pb({r, i});
adj[l+n].pb({r+n, i});
}
ans[i] = '1';
}
for (int i = 1; i <= n; i++) {
a[i] += a[i - 1];
if (a[i] == 1)
return !printf("impossible\n");
}
for (int i = 1; i <= n; i++) {
a[i] = max(0, min(1, a[i] - 2));
if (!a[i]) continue;
adj[i].pb({i - 1, -1});
adj[i + n].pb({i - 1 + n, -1});
}
for (int i = 0; i < n; i++) {
if (!ok(i)) continue;
for (int j: comp)
if (j != -1)
ans[j] = '0';
return !printf("%s\n", ans);
}
printf("impossible\n");
return 0;
}
