#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

const int M = 3e5+10, N = 1e3+10;

int n, m, k;

char s[N][N];

int l[M], l2[M], r[M], r2[M], e[M];

int c[26][N][N], d[26][N][N];

ll sum[N][N], val[26][N][N];

void upd(int c[N][N], int l, int r, int l2, int r2, int v) {

	c[l][r]+=v,c[l][r2+1]-=v,c[l2+1][r]-=v,c[l2+1][r2+1]+=v;

}

ll get(ll c[N][N], int id) {

	return c[l2[id]][r2[id]]-c[l[id]-1][r2[id]]-c[l2[id]][r[id]-1]+c[l[id]-1][r[id]-1];

}

int main() {

	scanf("%d%d%d", &n, &m, &k);

	REP(i,1,n) scanf("%s", s[i]+1);

	REP(i,1,n) REP(j,1,m) s[i][j]-='a';

	REP(i,1,k) {

		char cc;

		scanf("%d%d%d%d %c", l+i,r+i,l2+i,r2+i,&cc);

		e[i] = cc-'a';

		REP(t,0,25) upd(d[t],l[i],r[i],l2[i],r2[i],-1);

		REP(t,0,25) upd(c[t],l[i],r[i],l2[i],r2[i],abs(e[i]-t));

	}

	REP(i,1,n) REP(j,1,m) {

		REP(t,0,25) {

			d[t][i][j] += d[t][i-1][j]+d[t][i][j-1]-d[t][i-1][j-1];

			c[t][i][j] += c[t][i-1][j]+c[t][i][j-1]-c[t][i-1][j-1];

			val[t][i][j] = (ll)(k+d[t][i][j])*abs(t-s[i][j])+c[t][i][j];

		}

		sum[i][j] = val[s[i][j]][i][j]+sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];

		REP(t,0,25) val[t][i][j]+=val[t][i-1][j]+val[t][i][j-1]-val[t][i-1][j-1];

	}

	ll ans = 1e18;

	REP(i,1,k) ans=min(ans,sum[n][m]-get(sum,i)+get(val[e[i]],i));

	printf("%lld\n", ans);

}