codeforces 846D (二分+二维前缀和）Monitor

Recently Luba bought a monitor. Monitor is a rectangular matrix of size n × m. But then she started to notice that some pixels cease to work properly. Luba thinks that the monitor will become broken the first moment when it contains a square k × kconsisting entirely of broken pixels. She knows that q pixels are already broken, and for each of them she knows the moment when it stopped working. Help Luba to determine when the monitor became broken (or tell that it's still not broken even after all qpixels stopped working).

Input

The first line contains four integer numbers n, m, k, q (1 ≤ n, m ≤ 500, 1 ≤ k ≤ min(n, m), 0 ≤ q ≤ n·m) — the length and width of the monitor, the size of a rectangle such that the monitor is broken if there is a broken rectangle with this size, and the number of broken pixels.

Each of next q lines contain three integer numbers x_i, y_i, t_i (1 ≤ x_i ≤ n, 1 ≤ y_i ≤ m, 0 ≤ t ≤ 10⁹) — coordinates of i-th broken pixel (its row and column in matrix) and the moment it stopped working. Each pixel is listed at most once.

We consider that pixel is already broken at moment t_i.

Output

Print one number — the minimum moment the monitor became broken, or "-1" if it's still not broken after these q pixels stopped working.

Example

Input

2 3 2 52 1 82 2 81 2 11 3 42 3 2

Output

8

Input

3 3 2 51 2 22 2 12 3 53 2 102 1 100

Output

-1

考虑到坏了的就是坏了的，时间的符合性具有单调性，所以可以二分时间，找到符合的结果，在找

k*k的矩形里面用到了二维前缀和，sum记录 0,0到i，j的值，然后利用树状数组求对应矩形的做法就好，因为没有修改操作所以不需要后缀数组。

#include<bits/stdc++.h>using namespace std;int n,m,k;struct node{int x,y,t;}s[500000];int sum[510][510];int mian[510][510];int cmp(node a,node b){return a.t<b.t;}int check(int num){memset(mian,0,sizeof(mian));memset(sum,0,sizeof(sum));for(int i=1;i<=num;i++)mian[s[i].x][s[i].y]=1;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+mian[i][j];for(int i=k;i<=n;i++){for(int j=k;j<=m;j++)if(sum[i][j]-sum[i-k][j]-sum[i][j-k]+sum[i-k][j-k]==k*k)return 1;}return 0;}int main(){int q;scanf("%d%d%d%d",&n,&m,&k,&q);for(int i=1;i<=q;i++)scanf("%d%d%d",&s[i].x,&s[i].y,&s[i].t);sort(s+1,s+q+1,cmp);int l=1,r=q;int ans=-1;while(l<=r){int mid=(l+r)>>1;if(check(mid)) {ans=s[mid].t;r=mid-1;}else l=mid+1;}printf("%d\n",ans );}