#include<cstdio>

 #include<algorithm>

 #include<cstring>

 #include<cmath>

 #include<iostream>

 #include<cctype>

 #include<set>

 #include<vector>

 #include<queue>

 #include<map>

 using namespace std;

 typedef long long LL;

 inline int read() {

     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;

     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;

 }

 const int N = ;

 char a[N][N];

 int x[N], y[N], L[N], R[N], U[N][N], D[N][N], q1[N], q2[N], ans[N], sk[N];

 int Ans, n, m;

 void solve() {

     for (int i=; i<=n; ++i)

         for (int j=; j<=m; ++j)

             U[i][j] = a[i][j] == '.' ? U[i-][j] +  : ;

     for (int i=n; i>=; --i)

         for (int j=; j<=m; ++j)

             D[i][j] = a[i][j] == '.' ? D[i+][j] +  : ;

     for (int i=; i<=n; ++i) { // 单调栈

         U[i][] = U[i][m+] = -;

         for (int top=,j=; j<=m+; ++j) {

             while (top >  && U[i][j] < U[i][sk[top]]) R[sk[top]] = j, top--;

             sk[++top] = j;

         }

         for (int top=,j=m; j>=; --j) {

             while (top >  && U[i][j] < U[i][sk[top]]) L[sk[top]] = j, top --;

             sk[++top] = j;

         }

         for (int j=; j<=m; ++j)

             Ans = max(Ans, min(U[i][j], R[j] - L[j] - ));

     }

     /*memset(R, 0x3f, sizeof(R)); // 悬线法

     for (int i=1; i<=n; ++i) {

         int last = 0;

         for (int j=1; j<=m; ++j) {

             if (a[i][j] == '.') L[j] = max(L[j], last + 1);

             else last = j, L[j] = 0;

         }

         last = m + 1;

         for (int j=m; j>=1; --j) {

             if (a[i][j] == '.') R[j] = min(R[j], last - 1);

             else last = j, R[j] = m + 1;

         }

         for (int j=1; j<=m; ++j)

             Ans = max(Ans, min(U[i][j], R[j] - L[j] + 1));

     }*/

 }

 bool update(int x,int k) { // 判断能否组成k*k的方阵

     int L1 = , R1 = , L2 = , R2 = ;

     for (int j=; j<=m; ++j) {

         while (L1 <= R1 && j - q1[L1] +  > k) L1 ++;

         while (L1 <= R1 && U[x][q1[R1]] >= U[x][j]) R1 --;

         q1[++R1] = j;

         while (L2 <= R2 && j - q2[L2] +  > k) L2 ++;

         while (L2 <= R2 && D[x][q2[R2]] >= D[x][j]) R2 --;

         q2[++R2] = j;

         if (j >= k && U[x][q1[L1]] + D[x][q2[L2]] -  >= k) {

             Ans = k;

             return true;

         }

     }

     return false;

 }

 void Change(int x,int y) {

     a[x][y] = '.';

     for (int i=x; i<=n; ++i) U[i][y] = a[i][y] == '.' ? U[i-][y] +  : ;

     for (int i=x; i>=; --i) D[i][y] = a[i][y] == '.' ? D[i+][y] +  : ;

 }

 int main() {

     n = read(), m = read(); int Q = read();

     for (int i=; i<=n; ++i) scanf("%s",a[i]+);

     for (int i=; i<=Q; ++i) {

         x[i] = read(),y[i] = read();

         a[x[i]][y[i]] = 'X';

     }

     solve();

     for (int i=Q; i>=; --i) {

         ans[i] = Ans;

         Change(x[i], y[i]);

         while (update(x[i], Ans + ));

     }

     for (int i=; i<=Q; ++i) printf("%d\n",ans[i]);

     return ;

 }