#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define ld long double
#define pb push_back
#define mp make_pair

const int MAXN=205;
const ll inf = 1e18;
const ll mod=1000000007;

typedef pair<ll,ll> pi;

ll gc(ll a,ll b)
{
    if(b==0) return a;
    return gc(b,a%b);
}
ll power(ll x, ll y, ll p)
{
    ll res = 1;     // Initialize result

    x = x % p; // Update x if it is more than or
                // equal to p

    if (x == 0) return 0; // In case x is divisible by p;

    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;

        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res;
}

vector<bool>prime(1000005,true);
vector<ll>v;
void sieve(ll n)
{
    prime[1]=false;
   for (int p=2; p*p<=n; p++)
    {
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true)
        {
            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i=p*p; i<=n; i += p)
                prime[i] = false;
        }
    }

    for(int i=2;i<=n;i++)
    {
        if(prime[i]==true)
            v.push_back(i);
    }
}

int BIT[1000];
void update(int x, int delta)
{
    int n;
      for(; x <= n; x += x&-x)
        BIT[x] += delta;
}
int query(int x)
{
     int sum = 0;
     for(; x > 0; x -= x&-x)
        sum += BIT[x];
     return sum;
}


inline ll add(ll a,ll b) {return ((a%mod)+(b%mod))%mod;}
inline ll sub(ll a,ll b) {return ((a%mod)-(b%mod)+mod)%mod;}
inline ll mul(ll a,ll b) {return ((a%mod)*(b%mod))%mod;}


ll recur(ll idx,ll sum, vector<pair<ll,ll> >& v, vector<ll>& b, vector<ll>& pre, vector<ll>& dp)
{
    if(idx==b.size()) return 0;
    if(dp[idx]!=-1) return dp[idx];
    ll temp=INT_MAX;
    temp=min(temp,recur(idx+1,sum,v,b,pre,dp) + b[idx]);
    for(int i=0;i<v.size();i++)
    {
        ll x=v[i].first;
        ll y=v[i].second;
        if(idx+x-1>=b.size()) continue;
        ll calc= pre[idx+x-1]-pre[idx+y-1];
        temp=min(temp, recur(idx+x, sum,v,b,pre,dp)+ calc);
    }
    dp[idx]=temp;
    return temp;
}


ll find(ll u, vector<ll>& parent)
{
    if(u==parent[u]) return u;
    else return parent[u]=find(parent[u],parent);
}
 void unio(ll u, ll v, vector<ll>& parent, vector<ll>& ran)
{
    u=find(u,parent);
    v=find(v,parent);
    if(u==v) return;
    if(ran[u]>=ran[v])
    {
        parent[v]=u;
        ran[u]+=ran[v];
    }
    else{
        parent[u]=v;
        ran[v]+=ran[u];
    }
}
ll dfs(ll src, ll p, vector<ll>& a, vector<pair<ll,ll> > adj[], vector<ll>& dp)
{
    ll can1=0;
    ll can2=0;
    vector<pair<ll,ll> >:: iterator it;
    for(it=adj[src].begin();it!=adj[src].end();it++)
    {
        if(((*it).first==p)) continue;
        ll st=max(0LL,dfs((*it).first,src,a,adj,dp)- (*it).second);
        if(st>can1)
        {
            can2=can1;
            can1=st;
        }
        else if(st>can2)
        {
            can2=st;
        }
    }
    dp[src]=a[src]+ can1+can2;
    return (max(can1,can2)+ a[src]);
}
/*  FINDING LCS IN (NLOGN) COMPLEXITY
int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	vector<int> d(n+1, 1000000000);
	for (int i = 0; i < n; i++) {
		*lower_bound(d.begin(), d.end(), a[i]) = a[i];
	}
	for (int i = 0; i <= n; i++) {
		if (d[i] == 1000000000) {
			cout << i << endl;
			return 0;
		}
	}
*/
struct comp1{
    bool operator ()(pair<ll,ll>& temp1, pair<ll,ll>& temp2)
{

}
};
bool comp2(pair<ll,ll>& temp1, pair<ll,ll>& temp2)
{
    return ((temp1.first<temp2.first) || ((temp1.first==temp2.first) && (temp1.second<temp2.second)));
}

ll getinv(ll temp)
{
    return power(temp,mod-2,mod);

}
void computeLPSArray(string pat, ll M, vector<ll>&lps)
{
    // length of the previous longest prefix suffix
    int len = 0;

    lps[0] = 0; // lps[0] is always 0

    // the loop calculates lps[i] for i = 1 to M-1
    ll i = 1;
    while (i < M) {
        if (pat[i] == pat[len]) {
            len++;
            lps[i] = len;
            i++;
        }
        else // (pat[i] != pat[len])
        {
            // This is tricky. Consider the example.
            // AAACAAAA and i = 7. The idea is similar
            // to search step.
            if (len != 0) {
                len = lps[len - 1];

                // Also, note that we do not increment
                // i here
            }
            else // if (len == 0)
            {
                lps[i] = 0;
                i++;
            }
        }
    }
}


 struct node {
    ll sum, prefixsum, suffixsum, maxsum;
};

// array to store the segment tree
node tree[4 * 100005];

// function to build the tree
void build(vector<ll>& arr, ll low, ll high, ll index)
{
    // the leaf node
    if (low == high) {
        tree[index].sum = arr[low];
        tree[index].prefixsum = arr[low];
        tree[index].suffixsum = arr[low];
        tree[index].maxsum = arr[low];
    }
    else {
        ll mid = (low + high) / 2;

        // left subtree
        build(arr, low, mid, 2 * index + 1);

        // right subtree
        build(arr, mid + 1, high, 2 * index + 2);

        // parent node's sum is the summation
        // of left and right child
        tree[index].sum = tree[2 * index + 1].sum +
                          tree[2 * index + 2].sum;

        // parent node's prefix sum will be equivalent
        // to maximum of left child's prefix sum or left
        // child sum + right child prefix sum.
        tree[index].prefixsum =
                    max(tree[2 * index + 1].prefixsum,
                    tree[2 * index + 1].sum +
                    tree[2 * index + 2].prefixsum);

        // parent node's suffix sum will be equal to right
        // child suffix sum or right child sum + suffix
        // sum of left child
        tree[index].suffixsum =
                    max(tree[2 * index + 2].suffixsum,
                    tree[2 * index + 2].sum +
                    tree[2 * index + 1].suffixsum);

        // maxum will be the maximum of prefix, suffix of
        // parent or maximum of left child or right child
        // and summation of left child's suffix and right
        // child's prefix.
        tree[index].maxsum =
                    max(tree[index].prefixsum,
                    max(tree[index].suffixsum,
                    max(tree[2 * index + 1].maxsum,
                    max(tree[2 * index + 2].maxsum,
                    tree[2 * index + 1].suffixsum +
                    tree[2 * index + 2].prefixsum))));
    }
}


// fucntion to return answer to  every type-1 query
node query(vector<ll>& arr, ll index, ll low,
           ll high,  ll l, ll r)
{
    // initially all the values are INT_MIN
    node result;
    result.sum = result.prefixsum =
                 result.suffixsum =
                 result.maxsum = -inf;

    // range does not lies in this subtree
    if (r < low || high < l)
        return result;

    // complete overlap of range
    if (l <= low && high <= r)
        return tree[index];

    ll mid = (low + high) / 2;

    // right subtree
    if (l > mid)
        return query(arr, 2 * index + 2,
                     mid + 1, high, l, r);

    // left subtree
    if (r <= mid)
        return query(arr, 2 * index + 1,
                     low, mid, l, r);

    node left = query(arr, 2 * index + 1,
                      low, mid, l, r);
    node right = query(arr, 2 * index + 2,
                        mid + 1, high, l, r);

    // finding the maximum and returning it
    result.sum = left.sum + right.sum;
    result.prefixsum = max(left.prefixsum, left.sum +
                           right.prefixsum);

    result.suffixsum = max(right.suffixsum,
                       right.sum + left.suffixsum);
    result.maxsum = max(result.prefixsum,
                    max(result.suffixsum,
                    max(left.maxsum,
                    max(right.maxsum,
                    left.suffixsum + right.prefixsum))));

    return result;
}

int main()
{
     ll n,k,s;
     cin>>n>>k>>s;
     for(ll i=0;i<k;i++)
        cout<<s<<" ";
     for(ll i=0;i<n-k;i++)
     {
         if(s==1e9)
            cout<<s-1<<" ";
         else cout<<s+1<<" ";
     }
    return 0;
}


Measured memory_bytes: 1408