Codeforces 677D

D. Vanya and Treasure

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya is in the palace that can be represented as a grid n × m. Each room contains a single chest, an the room located in the i-th row and j-th columns contains the chest of type aij. Each chest of type x ≤ p - 1 contains a key that can open any chest of type x + 1, and all chests of type 1 are not locked. There is exactly one chest of type p and it contains a treasure.

Vanya starts in cell (1, 1) (top left corner). What is the minimum total distance Vanya has to walk in order to get the treasure? Consider the distance between cell (r1, c1) (the cell in the row r1 and column c1) and (r2, c2) is equal to |r1 - r2| + |c1 - c2|.

Input

The first line of the input contains three integers n, m and p (1 ≤ n, m ≤ 300, 1 ≤ p ≤ n·m) — the number of rows and columns in the table representing the palace and the number of different types of the chests, respectively.

Each of the following n lines contains m integers aij (1 ≤ aij ≤ p) — the types of the chests in corresponding rooms. It's guaranteed that for each x from 1 to p there is at least one chest of this type (that is, there exists a pair of r and c, such that arc = x). Also, it's guaranteed that there is exactly one chest of type p.

Output

Print one integer — the minimum possible total distance Vanya has to walk in order to get the treasure from the chest of type p.

Examples

input

Copy

3 4 3
2 1 1 1
1 1 1 1
2 1 1 3

output

Copy

5

input

Copy

3 3 9
1 3 5
8 9 7
4 6 2

output

Copy

22

input

Copy

3 4 12
1 2 3 4
8 7 6 5
9 10 11 12

output

Copy

11

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath> 
#define maxn 310
using namespace std;
struct node{
    int dis,pos;
}b[3050000],now;
int a[maxn][maxn];
int dp[maxn][maxn];
bool cmp(node a,node b){
    return a.dis<b.dis;
}
vector<node>mp[maxn*maxn];
int main(){
    int n,m,p;
    scanf("%d%d%d",&n,&m,&p);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            dp[i][j]=0x3f3f3f3f;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%d",&a[i][j]);
            if(a[i][j]==1){
                dp[i][j]=i+j-2;
                now.dis=dp[i][j];
            }
            now.pos=(i-1)*m+j;
            mp[a[i][j]].push_back(now);
        }
    }
    sort(mp[1].begin(),mp[1].end(),cmp);
    for(int i=2;i<=p;i++){
        int sz1=mp[i].size();
        int sz2=mp[i-1].size();
        for(int j=0;j<sz1;j++){
            int x=mp[i][j].pos/m+1;
            int y=mp[i][j].pos%m;
            if(y==0)y=m,x--;
            for(int k=0;k<sz2&&k<1000;k++){
                int xx=mp[i-1][k].pos/m+1;
                int yy=mp[i-1][k].pos%m;
                if(yy==0)yy=m,xx--;
                dp[x][y]=min(dp[x][y],dp[xx][yy]+abs(x-xx)+abs(y-yy));
            }
            mp[i][j].dis=dp[x][y];
        }
        sort(mp[i].begin(),mp[i].end(),cmp);
    }
    int ans=0x3f3f3f3f;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            if(a[i][j]==p)
            ans=min(dp[i][j],ans);
        }
    printf("%d
",ans);
    return 0;
}