Illumination

You inherited a haunted house. Its floor plan is an $n$-by-$n$ square grid with $l$ lamps in fixedlocations and no interior walls. Each lamp can either illuminate its row or its column, but not bothsimultaneously. The illumination of each lamp extends by $r$ squares in both directions, so a lampunobstructed by an exterior wall of the house can illuminate as many as $2r + 1$ squares.

If a square is illuminated by more than one lamp in its row, or by more than one lamp in its column,the resulting bright spot will scare away ghosts forever, diminishing the value of your property. Isit possible for all lamps to illuminate a row or column, without scaring any ghosts? Note that asquare illuminated by two lamps, one in its row and the other in its column, will not scare awaythe ghosts.

Input

The first line of input contains three positive integers, $n$, $r$ and $l$$(1 ≤ n, r, l ≤ 1,000)$.

Each of the next $l$ lines contains two positive integers $r_i$ and $c_i$ $(1 ≤ r_i, c_i ≤ n)$, indicating thatthere is a lamp in row ri and column $c_i$.

It is guaranteed that all lamps are in distinct locations.

Output

Print, on a single line, YES if it is possible to illuminate all lamps as stated above; otherwise, print NO.

樣例輸入1
3 2 5
1 1
1 3
3 1
3 3
2 2
樣例輸出1
YES
樣例輸入2
3 2 6
1 1
1 2
1 3
3 1
3 2
3 3
樣例輸出2
NO

對於處於相同行的燈$i$，$j$，如果分別位於$c_i$列和$c_j$列，且兩個燈在行方向上能夠相互影響到，即$|c_i-c_j|≤2·r$，則若燈$i$在行方向上，則燈$j$在列方向上；對於處於同列的燈同理。
這就是圖論上的2-SAT問題。

#include<bits/stdc++.h>

#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scahf("%c",&a);
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pc(a) putchar(a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define pii pair<int,int>
#define mii unordered_map<int,int>
#define msi unordered_map<string,int>
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define pr(x) cout<<#x<<": "<<x<<endl
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 2e3 + 10, M = 1e6 + 10;
vi col[N], row[N];
map<pii, int> id;
int head[N], ver[M << 1], Next[M << 1], tot;
stack<int> s;
bool ins[N];
int c[N], dfn[N], low[N], cnt, num;
int n, k, m;

inline void add(int x, int y) {
    ver[++tot] = y;
    Next[tot] = head[x];
    head[x] = tot;
}

inline void build() {
    repi(i, 1, n) if (row[i].size() >= 2) {
            int l = 0, r = 0;
            repi(j, 0, row[i].size() - 1) {
                while (row[i][j] - row[i][l] > 2 * k)l++;
                while (r + 1 < row[i].size() && row[i][r + 1] - row[i][j] <= 2 * k) r++;
                repi(t, l, r) if (t != j)add(id[{i, row[i][j]}], id[{i, row[i][t]}] + m);
            }
        }
    repi(i, 1, n) if (col[i].size() >= 2) {
            int l = 0, r = 0;
            repi(j, 0, col[i].size() - 1) {
                while (col[i][j] - col[i][l] > 2 * k)l++;
                while (r + 1 < col[i].size() && col[i][r + 1] - col[i][j] <= 2 * k)r++;
                repi(t, l, r) if (t != j)add(id[{col[i][t], i}] + m, id[{col[i][j], i}]);
            }
        }
}

void tarjain(int x) {
    dfn[x] = low[x] = ++num;
    s.push(x);
    ins[x] = true;
    for (int i = head[x]; i; i = Next[i])
        if (!dfn[ver[i]]) {
            tarjain(ver[i]);
            low[x] = min(low[x], low[ver[i]]);
        } else if (ins[ver[i]])
            low[x] = min(low[x], dfn[ver[i]]);
    if (dfn[x] == low[x]) {
        cnt++;
        int y;
        do {
            y = s.top();
            s.pop();
            ins[y] = false;
            c[y] = cnt;
        } while (x != y);
    }
}

inline void solve() {
    bool flag = true;
    for (int i = 1; i <= m; i++)
        if (c[i] == c[i + m]) {
            flag = false;
            break;
        }
    puts(flag ? "YES" : "NO");
}

int main() {
    n = qr(), k = qr(), m = qr();
    repi(i, 1, m) {
        int x = qr(), y = qr();
        id[{x, y}] = i;
        row[x].pb(y), col[y].pb(x);
    }
    repi(i, 1, n) {
        sort(row[i].begin(), row[i].end());
        sort(col[i].begin(), col[i].end());
    }
    build();
    repi(i, 1, 2 * m)if (!dfn[i])tarjain(i);
    solve();
    return 0;
}