#include <bits/stdc++.h>
using namespace std;

int N, M, K;
vector<pair<int, int>> ans;

void print_no() {
    cout << "No\n";
}

void print_yes() {
    cout << "Yes\n";
    for (auto [u, v] : ans) {
        cout << u << ' ' << v << '\n';
    }
}

int main() {
    cin >> N >> M >> K;

    int T = N * (N - 1) / 2;

    if (K == 1) {
        if (M == T) {
            print_no();
            return 0;
        }

        for (int u = 1; u <= N; ++u) {
            for (int v = u + 1; v <= N; ++v) {
                if (u == 1 && v == N) continue;
                if ((int)ans.size() < M) {
                    ans.push_back({u, v});
                }
            }
        }

        print_yes();
        return 0;
    }

    // K >= 2.
    int max_removed = T - K;

    // WRONG:
    // It checks only the upper bound of M.
    // It forgets the lower bound needed to remove shortcuts.
    if (M > max_removed) {
        print_no();
        return 0;
    }

    // Use the path 1 - 2 - ... - K - N.
    vector<int> path;
    path.push_back(1);
    for (int v = 2; v <= K; ++v) {
        path.push_back(v);
    }
    path.push_back(N);

    vector<vector<bool>> is_path_edge(N + 1, vector<bool>(N + 1, false));
    for (int i = 0; i + 1 < (int)path.size(); ++i) {
        int a = path[i];
        int b = path[i + 1];
        if (a > b) swap(a, b);
        is_path_edge[a][b] = true;
    }

    // Intended distance layer (BFS tree) of each vertex.
    vector<int> layer(N + 1, -1);
    layer[1] = 0;
    for (int i = 1; i <= K - 1; ++i) {
        layer[i + 1] = i;
    }
    layer[N] = K;

    // Make extra vertices belong to layer 1.
    for (int v = K + 1; v <= N - 1; ++v) {
        layer[v] = 1;
    }

    vector<vector<bool>> removed(N + 1, vector<bool>(N + 1, false));

    // WRONG:
    // Instead of first removing all shortcut edges,
    // this code just removes arbitrary non-path edges.
    // Therefore, many shortcut edges may remain.
    for (int u = 1; u <= N && (int)ans.size() < M; ++u) {
        for (int v = u + 1; v <= N && (int)ans.size() < M; ++v) {
            if (is_path_edge[u][v]) continue;

            removed[u][v] = true;
            ans.push_back({u, v});
        }
    }

    print_yes();
    return 0;
}