#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
#include <string>
#include <functional>
#include <vector>
#include <numeric>
#include <deque>
#include <utility>
#include <bitset>
#include <iostream>
using namespace std;
typedef long long lint;
typedef long double llf;
typedef pair<int, int> pi;
char str[2005][2005];
bool vis[2005][2005];
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
int n, m;
int arr[2005][2005], dp[2005][2005];
vector<pi> pnt;
queue<int> qx, qy;
void bfs(int x, int y){
qx.push(x);
qy.push(y);
vis[x][y] = 1;
while(!qx.empty()){
int x = qx.front();
int y = qy.front();
qx.pop(), qy.pop();
pnt.push_back(pi(y, str[x][y] - '0'));
for(int i=0; i<4; i++){
if(x + dx[i] < 0 || x + dx[i] >= n || y + dy[i] < 0 || y + dy[i] >= m){
continue;
}
if(str[x+dx[i]][y+dy[i]] == '.') continue;
if(vis[x + dx[i]][y + dy[i]]) continue;
vis[x + dx[i]][y + dy[i]] = 1;
qx.push(x + dx[i]);
qy.push(y + dy[i]);
}
}
}
inline int cost(int i, int j){
return arr[j][j] - arr[i-1][j] - arr[j][i-1] + arr[i-1][i-1];
}
void solve(int *pdp, int *dp, int s, int e, int ps, int pe){
if(ps > pe) return;
int pm = (ps + pe) / 2;
int opt = -1;
for(int i=s; i<=min(e, pm-1); i++){
if(dp[pm] > pdp[i] + cost(i+1, pm-1)){
dp[pm] = pdp[i] + cost(i+1, pm-1);
opt = i;
}
}
solve(pdp, dp, s, opt, ps, pm-1);
solve(pdp, dp, opt, e, pm+1, pe);
}
int main(){
scanf("%d %d",&n,&m);
for(int i=0; i<n; i++){
scanf("%s",str[i]);
}
for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
if(!vis[i][j] && str[i][j] != '.'){
pnt.clear();
bfs(i, j);
int ret = 0, sy = 1e9, ey = 0;
for(auto &i : pnt){
ret += i.second;
sy = min(sy, i.first);
ey = max(ey, i.first);
}
arr[sy+1][ey+1] += ret;
}
}
}
for(int i=1; i<=m+1; i++){
for(int j=1; j<=m+1; j++){
arr[i][j] += arr[i][j-1] + arr[i-1][j] - arr[i-1][j-1];
}
}
memset(dp, 0x3f, sizeof(dp));
dp[0][0] = 0;
solve(dp[0], dp[1], 0, m, 1, m+1);
for(int i=2; i<=m+1; i++){
solve(dp[i-1], dp[i], 0, m+1, 0, m+1);
printf("%d\n", arr[m][m] - dp[i][m+1]);
}
}
