Problem: The sales department of Japanese Ancient Giant Corp. is visiting a hot spring resort for their recreational trip. For deepening their friendships, they are staying in one large room of a Japanese-style hotel called a ryokan.

In the ryokan, people sleep in Japanese-style beds called futons. They all have put their futons on the floor just as they like. Now they are ready for sleeping but they have one concern: they don’t like to go into their futons with their legs toward heads — this is regarded as a bad custom in Japanese tradition. However, it is not obvious whether they can follow a good custom. You are requested to write a program answering their question, as a talented programmer.

Here let's model the situation. The room is considered to be a grid on an xy-plane. As usual, x-axis points toward right and y-axis points toward up. Each futon occupies two adjacent cells. People put their pillows on either of the two cells. Their heads come to the pillows; their foots come to the other cells. If the cell of some person's foot becomes adjacent to the cell of another person's head, regardless their directions, then it is considered as a bad case. Otherwise people are all right.



Input

The input is a sequence of datasets. Each dataset is given in the following format:

n
x1 y1 dir1
...
xn yn dirn


n is the number of futons (1 ≤ n ≤ 20,000); (xi, yi) denotes the coordinates of the left-bottom corner of the i-th futon; diri is either 'x' or 'y' and denotes the direction of the i-th futon, where 'x' means the futon is put horizontally and 'y' means vertically. All coordinate values are non-negative integers not greater than 109 .

It is guaranteed that no two futons in the input overlap each other.

The input is terminated by a line with a single zero. This is not part of any dataset and thus should not be processed.

Output

For each dataset, print "Yes" in a line if it is possible to avoid a bad case, or "No" otherwise.

Example

Input

4
0 0 x
2 0 x
0 1 x
2 1 x
4
1 0 x
0 1 x
2 1 x
1 2 x
4
0 0 x
2 0 y
0 1 y
1 2 x
0


Output

Yes
No
Yes

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define int long long   // <-----!!!!!!!!!!!!!!!!!!!

#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n)-1;i>=0;i--)
#define rrep2(i,a,b) for (int i=(a)-1;i>=b;i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define printV(_v) for(auto _x:_v){cout<<_x<<" ";}cout<<endl
#define printVS(_vs) for(auto _x : _vs){cout << _x << endl;}
#define printVV(_vv) for(auto _v:_vv){for(auto _x:_v){cout<<_x<<" ";}cout<<endl;}
#define printP(_p) cout << _p.first << " " << _p.second << endl
#define printVP(_vp) for(auto _p : _vp) printP(_p);

typedef long long ll;
typedef pair<int, int> Pii;
typedef tuple<int, int, int> TUPLE;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<vector<int>> Graph;
const int inf = 1e9;
const int mod = 1e9 + 7;

struct Point {
    int x, y, id;
    Point(){}
    Point(int _x, int _y, int _id) : x(_x), y(_y), id(_id) {}
};

bool cmp_Yx(const Point& p, const Point& q) {
    if (p.y != q.y) return p.y < q.y;
    if (p.x != q.x) return p.x < q.x;
    assert(false);
}

bool cmp_Xy(const Point& p, const Point& q) {
    if (p.x != q.x) return p.x < q.x;
    if (p.y != q.y) return p.y < q.y;
    assert(false);
}

bool adjecent(const Point& p, const Point& q) {
    if (p.x == q.x && abs(p.y - q.y) == 1) return true;
    if (p.y == q.y && abs(p.x - q.x) == 1) return true;
    return false;
}

// undirected
void addEdge(Graph& G, int x, int y) {
    G[x].emplace_back(y);
    G[y].emplace_back(x);
}

void solve(int n) {
    vector<Point> p;
    rep(i, n) {
        int x, y;
        char dir;
        cin >> x >> y >> dir;
        p.emplace_back(x, y, 2 * i);
        if (dir == 'x') {
            p.emplace_back(x + 1, y, 2 * i + 1);
        } else {
            p.emplace_back(x, y + 1, 2 * i + 1);
        }
    }

    Graph G(2 * n);

    // horizontal edge
    sort(all(p), cmp_Yx);
    rep(i, 2 * n - 1) {
        if (p[i].y == p[i + 1].y && p[i + 1].x - p[i].x == 1) {
            addEdge(G, p[i].id, p[i + 1].id);
        }
    }

    // vertical edge
    sort(all(p), cmp_Xy);
    rep(i, 2 * n - 1) {
        if (p[i].x == p[i + 1].x && p[i + 1].y - p[i].y == 1) {
            addEdge(G, p[i].id, p[i + 1].id);
        }
    }

    // printVV(G);

    vector<int> color(2 * n, -1);

    std::function<bool(int,int)> dfs = [&](int i, int c) {
        // cout << i << " " << c << endl;
        color[i] = c;

        for (auto j : G[i]) {
            if (color[j] == -1) continue;
            if (i / 2 == j / 2 && color[i] == color[j]) {
                // same futon but same color
                return false;
            }
            if (i / 2 != j / 2 && color[i] != color[j]) {
                // adjecent cells in different futons but different color
                return false;
            }
        }

        for (auto j : G[i]) {
            if (color[j] != -1) continue;
            if (i / 2 == j / 2) {
                // same futon
                if (!dfs(j, c^1)) {
                    return false;
                }
            } else {
                // other futon
                if (!dfs(j, c)) {
                    return false;
                }
            }
        }

        return true;
    };

    rep(i, 2 * n) {
        // cout << "i = " << i << ":" << endl;
        if (color[i] == -1 && !dfs(i, 0)) {
            cout << "No" << endl;
            return;
        }
    }

    // printV(color);

    cout << "Yes" << endl;
    // cout << endl;
}

signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    int n;
    while (cin >> n, n) {
        solve(n);
    }
}