#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
struct node
{
    int sign, dep, id;
};
vector<node> edge[maxn];
int dep[maxn];
int sign[maxn];
vector<pair<int, int>> ans;

void dfs(int now, int fa)
{
    dep[now] = dep[fa] + 1;
    for (int i = 0; i < edge[now].size(); ++i)
    {
        int id = edge[now][i].id;
        if (id == fa)
            continue;
        dfs(id, now);
    }
}

node dfs1(int now, int fa)
{
    vector<pair<int, int>> tmp[4]; // 待处理序列 tmp[sign] = {<dep,id>}
    node retu = node{0, 0, 0};     // 当层返回值，id=-1表示无解，0表示没有点
    for (node a : edge[now])
    {
        if (a.id == fa)
            continue;
        node rec = dfs1(a.id, now); // return仅能返回一个点的值
        if (rec.sign == -1)         // 返回-1说明无解
            return node{-1, -1, -1};
        else if (rec.sign > 0)                          // 返回1,2,3说明有返回点
            tmp[rec.sign].push_back({rec.dep, rec.id}); // 加入待处理序列
    }                                                   // 所有儿子节点情况已经添加到tmp中
    if (sign[now] != 0)                                 // 自己拥有标记，加入匹配序列
        tmp[sign[now]].push_back({dep[now], now});
    for (int i = 1; i <= 3; ++i)
        sort(tmp[i].begin(), tmp[i].end(), greater<pair<int, int>>()); // 按照dep降序排序
    if (tmp[3].size())
    {
        int jishu = 1;
        for (int i = 1; i < tmp[3].size(); ++i)
        {
            if (tmp[3][i].first != tmp[3][i - 1].first) // 存在tong无法匹配
            {
                if (jishu % 2) // 奇数个tong
                {
                    if (retu.sign != 0)                                        // 存在待返回值
                        return node{-1, -1, -1};                               // 直接返回无解
                    retu = node{3, tmp[3][i - 1].first, tmp[3][i - 1].second}; // 否则加入返回值
                }
                for (int j = i - jishu; j < i - 1; j += 2)
                    ans.push_back({tmp[3][j].second, tmp[3][j + 1].second});
                jishu = 1;
            }
            else
                jishu++;
        }
        for (int i = tmp[3].size() - jishu; i < tmp[3].size() - 1; i += 2) // 处理最后一段tong
            ans.push_back({tmp[3][i].second, tmp[3][i + 1].second});
        if (jishu % 2) // 奇数个tong,最后一位多余
        {
            if (retu.sign != 0)
                return node{-1, -1, -1};
            retu = node{3, tmp[3][tmp[3].size() - 1].first, tmp[3][tmp[3].size() - 1].second};
        }
    }
    if (tmp[1].size() == tmp[2].size()) // chang,duan长度相同
    {
        for (int i = 0; i < tmp[1].size(); i++)
        {
            if (tmp[1][i].first <= tmp[2][i].first)
                return (node){-1, -1, -1};
            ans.push_back({tmp[1][i].second, tmp[2][i].second});
        }
    }
    else if (tmp[1].size() > tmp[2].size())
    {
        int zz = tmp[1].size() - 1;                  // zz表示chang的指针
        for (int i = tmp[2].size() - 1; i >= 0; i--) // i表示duan的指针
        {
            while (zz >= 0 && tmp[1][zz].first <= tmp[2][i].first) // 若长指针深度小于短指针深度
            {
                if (retu.sign != 0)
                    return node{-1, -1, -1};
                retu = node{1, tmp[1][zz].first, tmp[1][zz].second};
                zz--;
            }
            if (zz >= 0 && tmp[1][zz].first > tmp[2][i].first) // 长指针深度大于短指针深度
            {
                ans.push_back({tmp[1][zz].second, tmp[2][i].second}); // 匹配
                zz--;                                                 // 移位
            }
            else
                return node{-1, -1, -1};
        }
        while (zz >= 0) // 如果duan匹配完了，但是chang还有剩余
        {
            if (retu.sign != 0)
                return node{-1, -1, -1};
            retu = node{1, tmp[1][zz].first, tmp[1][zz].second};
            zz--; // 剩一个且之前没剩
        }
    }
    else if (tmp[1].size() < tmp[2].size()) // 同理
    {
        int zz = 0;
        for (int i = 0; i < tmp[1].size(); i++)
        {
            while (zz < tmp[2].size() && tmp[1][i].first <= tmp[2][zz].first)
            {
                if (retu.sign != 0)
                    return (node){-1, -1, -1};
                retu = (node){2, tmp[2][zz].first, tmp[2][zz].second}, zz++;
            }
            if (zz < tmp[2].size() && tmp[1][i].first > tmp[2][zz].first)
            {
                ans.push_back({tmp[1][i].second, tmp[2][zz].second});
                zz++;
            }
            else
                return node{-1, -1, -1};
        }
        while (zz < tmp[2].size())
        {
            if (retu.sign != 0)
                return node{-1, -1, -1};
            retu = node{2, tmp[2][zz].first, tmp[2][zz].second};
            zz++;
        }
    }
    return retu;
}
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i < n; ++i)
    {
        int a, p, si = 0;
        string t;
        cin >> a >> p >> t;
        if (t == "-")
            si = 0;
        else if (t == "Chang")
            si = 1;
        else if (t == "Duan")
            si = 2;
        else if (t == "Tong")
            si = 3;
        edge[p].push_back(node{si, 0, a});
        sign[a] = si;
    }
    dep[1] = 1;
    dfs(1, 0);
    node finish = dfs1(1, 0);
    if (finish.sign == 0)
    {
        cout << "YES" << endl;
        for (int i = 0; i < ans.size(); i++)
        {
            if (i != ans.size() - 1)
                cout << ans[i].first << " " << ans[i].second << endl;
            else
                cout << ans[i].first << " " << ans[i].second;
        }
    }
    else
        cout << "NO";
    return 0;
}