#include<bits/stdc++.h>
using namespace std;

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={0,1,0,-1,1,1,-1,-1,0};
const ll dx[9]={1,0,-1,0,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
/*struct pack{
    ll idx,l,m,r;
    pack(ll a):idx(a),l(0),m(0),r(0){}
    const bool operator<(const pack b) const{
    }
}*/
void output(vl ans){
    vector<string> g(3);
    rep(i,3)g[i]="...";
    rep(i,3){
        rep(j,ans[i]){
            g[j][i]='#';
        }
    }
    rep(i,3)cout << g[i] << endl;
}
int main(){
    ll h,w,n;cin >> h >> w >> n;
    vector<string> g(h);rep(i,h)cin >> g[i];
    vvl ans(n,vl(3));
    set<pl> need;
    vvpl use(w);
    rep(i,n){
        ll c;cin >> c;
        need.insert({c,i});
        rep(k,3)use[k+c].emplace_back(make_pair(i,k));
    }
    rep(i,w){
        ll cnt=0;
        rep(j,h)if(g[j][i]=='#')cnt++;
        while(cnt){
            auto p=need.lower_bound(make_pair(i-2,-INF));
            if(p==need.end())break;
            if(p->first>i)break;
            cnt--;
            ans[p->second][i-p->first]++;
            need.erase(p);
        }
        for(auto p:use[i]){
            if(cnt==0)break;
            ll f=min(cnt,3-ans[p.first][p.second]);
            cnt-=f;
            ans[p.first][p.second]+=f;
        }
    }
    rep(i,n)output(ans[i]);
    /*rep(i,n){
        rep(j,3)cout << ans[i][j] <<" ";cout << endl;
    }*/
}