codeforces—— 869E —— The Untended Antiquity

E. The Untended Antiquity

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Adieu l'ami.

Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around the abandoned Eikou Cram School building, Oshino's makeshift residence.

The space is represented by a rectangular grid of n × m cells, arranged into n rows and m columns. The c-th cell in the r-th row is denoted by (r, c).

Oshino places and removes barriers around rectangular areas of cells. Specifically, an action denoted by "1 r1 c1 r2 c2" means Oshino's placing barriers around a rectangle with two corners being (r1, c1) and (r2, c2) and sides parallel to squares sides. Similarly, "2 r1 c1 r2 c2" means Oshino's removing barriers around the rectangle. Oshino ensures that no barriers staying on the ground share any common points, nor do they intersect with boundaries of the n × m area.

Sometimes Koyomi tries to walk from one cell to another carefully without striding over barriers, in order to avoid damaging various items on the ground. "3 r1 c1 r2 c2" means that Koyomi tries to walk from (r1, c1) to (r2, c2) without crossing barriers.

And you're here to tell Koyomi the feasibility of each of his attempts.

Input

The first line of input contains three space-separated integers n, m and q (1 ≤ n, m ≤ 2 500, 1 ≤ q ≤ 100 000) — the number of rows and columns in the grid, and the total number of Oshino and Koyomi's actions, respectively.

The following q lines each describes an action, containing five space-separated integers t, r1, c1, r2, c2 (1 ≤ t ≤ 3, 1 ≤ r1, r2 ≤ n, 1 ≤ c1, c2 ≤ m) — the type and two coordinates of an action. Additionally, the following holds depending on the value of t:

• If t = 1: 2 ≤ r1 ≤ r2 ≤ n - 1, 2 ≤ c1 ≤ c2 ≤ m - 1;
• If t = 2: 2 ≤ r1 ≤ r2 ≤ n - 1, 2 ≤ c1 ≤ c2 ≤ m - 1, the specified group of barriers exist on the ground before the removal.
• If t = 3: no extra restrictions.

Output

For each of Koyomi's attempts (actions with t = 3), output one line — containing "Yes" (without quotes) if it's feasible, and "No" (without quotes) otherwise.

Examples

input

5 6 51 2 2 4 51 3 3 3 33 4 4 1 12 2 2 4 53 1 1 4 4

output

NoYes

input

2500 2500 81 549 1279 1263 21891 303 795 1888 24321 2227 622 2418 11613 771 2492 1335 14331 2017 2100 2408 21603 48 60 798 7291 347 708 1868 7923 1940 2080 377 1546

output

NoYesNo

Note

For the first example, the situations of Koyomi's actions are illustrated below.

方案1,添加障碍矩形   (line((x1,y1),(x2,y2)),为主对角线)方案2 删除障碍矩形   (line((x1,y1),(x2,y2)),为主对角线)方案3  查询能否在不触及障碍的情况下,把两个点连接起来

做一个结构体存储每一条内容的信息,set保留未删除的障碍物...通过查询两个点是否出现 一个在障碍外,一个在障碍里 来判断结果方法比较慢,比较暴力....

#include <cstring>#include <queue>#include <cstdio>#include <algorithm>#include <string>#include <map>#include <set>#include <vector>#include <iostream>#include <cmath>using namespace std;struct AP {    int x1,x2,y1,y2;    bool operator <(const AP &a)const {        if(x1!=a.x1) return x1<a.x1;        if(x2!=a.x2) return x2<a.x2;        if(y1!=a.y1) return y1<a.y1;        return y2<a.y2;    }    AP & operator = (const AP & a) {        x1=a.x1;        x2=a.x2;        y1=a.y1;        y2=a.y2;        return *this;    }};bool func1(int x,int y,AP pla) {    if(x>=min(pla.x1,pla.x2)&&x<=max(pla.x1,pla.x2))        if(y>=min(pla.y1,pla.y2)&&y<=max(pla.y1,pla.y2))        return true;    return false;}bool func2(AP val,AP pla) {    int ans=0;    ans+=func1(val.x1,val.y1,pla);    ans+=func1(val.x2,val.y2,pla);    if(ans==1) return false;    return true;}int main() {    int x,y,z;    while(~scanf("%d%d%d",&x,&y,&z)) {        int vju,x1,y1,x2,y2;        set<AP>se1;        while(z--) {            AP t;            scanf("%d%d%d%d%d",&vju,&t.x1,&t.y1,&t.x2,&t.y2);            if(vju==1) se1.insert(t);            if(vju==2) se1.erase(se1.find(t));            if(vju==3) {                set<AP>::iterator it;                for(it=se1.begin(); it!=se1.end(); it++)                    if(!func2(t,(*it))) {                        vju=0;                        break;                    }                if(vju) printf("Yes\n");                else printf("No\n");            }        }    }    return 0;}