#include <iostream>
#include <sstream>
#include <map>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;

#define all(c) (c).begin(), (c).end()
#define loop(i,a,b) for(int i=(a);i<int(b);i++)
#define rep(i,b) loop(i,0,b)

// Verify : http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2429
struct PrimalDual {
    typedef int Capacity;
    typedef int Cost;
    struct Edge;
    int n;
    vector<vector<Edge> > g;
    enum : Cost {
        inf = 1<<29,
        MAX_V = 1000000
    };

    struct Edge {
        int dst;
        Capacity cap;
        Capacity cap_orig;
        Cost cost;
        int revEdge;
        bool isRev;
        Edge(int dst, Capacity cap, Cost cost, int revEdge, bool isRev)
            :dst(dst), cap(cap), cap_orig(cap), cost(cost), revEdge(revEdge), isRev(isRev) {}
    };

    PrimalDual(int n_)
        : n(n_), g(vector<vector<Edge> >(n_)){}

    void add_edge(int src, int dst, Capacity cap, Cost cost) {
        g[src].push_back(Edge(dst, cap, cost, g[dst].size(), false));
        g[dst].push_back(Edge(src, 0, -cost, g[src].size() - 1, true));
    }

    int min_cost_flow(int s, int t, int f) {
        int res = 0;
        // vector<Cost> h(g.size()), dist(g.size());
        // vector<int> prevv(g.size()), preve(g.size());
        static Cost h[MAX_V];
        static Cost dist[MAX_V];
        static Cost prevv[MAX_V];
        static Cost preve[MAX_V];
        rep(i,g.size())h[i] = 0;
        while (f > 0) {
            typedef pair<Cost,int> pcv;
            priority_queue<pcv, vector<pcv>, greater<pcv> > q;
            rep(i,g.size()) dist[i] = inf;
            dist[s] = 0;
            q.push(pcv(0, s));
            while (q.size()) {
                pcv p = q.top(); q.pop();
                int v = p.second;
                if (dist[v] < p.first) continue;
                rep(i,g[v].size()){
                    Edge &e = g[v][i];
                    if (e.cap > 0 && dist[e.dst] > dist[v] + e.cost + h[v] - h[e.dst]) {
                        dist[e.dst] = dist[v] + e.cost + h[v] - h[e.dst];
                        prevv[e.dst] = v;
                        preve[e.dst] = i;
                        q.push(pcv(dist[e.dst], e.dst));
                    }
                }
            }
            if (dist[t] == inf) {
                return -1;
            }
            rep(v,n) h[v] += dist[v];
            // s-t 間最短路に沿って目一杯流す
            int d = f;
            for (int v = t; v != s; v = prevv[v]) {
                d = min(d, g[prevv[v]][preve[v]].cap);
            }
            f -= d;
            res += d * h[t];
            for (int v = t; v != s; v = prevv[v]) {
                Edge &e = g[prevv[v]][preve[v]];
                e.cap -= d;
                g[v][e.revEdge].cap += d;
            }
        }
        return res;
    }

    // 流れたフロー=元々の容量-現在の容量を表示
    void view(){
        rep(i,g.size()){
            rep(j,g[i].size())if(!g[i][j].isRev){
                Edge& e = g[i][j];
                printf("%3d->%3d (flow:%d)\n", i, e.dst, e.cap_orig - e.cap);
            }
        }
    }
};

int W[111][111], E[111][111];
char f[111][111];
int n;
int N;

void solve(){
    int C=0;
    rep(i,n)rep(j,n){
        if(f[i][j]=='o')C+=E[i][j];
    }
    PrimalDual mcf(n*2+2);
    int s=n*2, t=n*2+1;
    rep(i,n){
        mcf.add_edge(s,i,1,0);
        mcf.add_edge(i+n,t,1,0);
    }
    rep(i,n)rep(j,n){
        if(f[i][j]=='o'){
            mcf.add_edge(i,j+n,1,-E[i][j]);
        }else{
            mcf.add_edge(i,j+n,1,W[i][j]);
        }
    }
    printf("%d\n",C+mcf.min_cost_flow(s,t,n));
    static int ans[10000][3];
    int cnt=0;

    rep(i,mcf.g.size()){
        rep(j,mcf.g[i].size()){
            auto & e=mcf.g[i][j];
            if(e.isRev || e.cost==0) continue;
            int r = i, c = e.dst-n;
            if(f[r][c]=='o' && e.cap!=0){
                cnt++;
                ans[cnt][0]=r+1;
                ans[cnt][1]=c+1;
                ans[cnt][2]=0;
            }else if(f[r][c]=='.' && e.cap==0){
                cnt++;
                ans[cnt][0]=r+1;
                ans[cnt][1]=c+1;
                ans[cnt][2]=1;
            }
        }
    }
    printf("%d\n",cnt);
    rep(i,cnt){
        printf("%d %d %s\n",ans[i][0],ans[i][1],ans[i][2]==0 ? "erase" : "write");
    }
}

int main(){
    while(cin >> n){
        N=n*2+2;
        rep(i,n)rep(j,n)cin>>W[i][j];
        rep(i,n)rep(j,n)cin>>E[i][j];
        rep(i,n)cin>>f[i];
        solve();
    }
}

Measured memory_bytes: 8028