N, M, K = map(int, input().split())

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

ans = []


def print_no():
    print("No")


def print_yes():
    print("Yes")
    for u, v in ans:
        print(u, v)


if K == 1:
    # Distance is 1 iff the edge (1, N) remains.
    if M == T:
        print_no()
        exit()

    for u in range(1, N + 1):
        for v in range(u + 1, N + 1):
            if u == 1 and v == N:
                continue
            if len(ans) < M:
                ans.append((u, v))

    print_yes()
    exit()


# K >= 2.
min_removed = K * (K - 1) // 2 + (N - K - 1) * (K - 2)
max_removed = T - K  # at least K edges of a shortest path must remain.

if not (min_removed <= M <= max_removed):
    print_no()
    exit()

# Use the path 1 - 2 - ... - K - N.
path = []
path.append(1)
for v in range(2, K + 1):
    path.append(v)
path.append(N)

is_path_edge = [[False] * (N + 1) for _ in range(N + 1)]
for i in range(len(path) - 1):
    a = path[i]
    b = path[i + 1]
    if a > b:
        a, b = b, a
    is_path_edge[a][b] = True

# Intended distance layer (BFS tree) of each vertex.
layer = [-1] * (N + 1)
layer[1] = 0
for i in range(1, K):
    layer[i + 1] = i
layer[N] = K

# Make extra vertices belong to layer 1.
for v in range(K + 1, N):
    layer[v] = 1

removed = [[False] * (N + 1) for _ in range(N + 1)]

# First remove all edges whose layer difference is at least 2 (Since it makes a shortcut).
# This is the minimum-removal construction.
for u in range(1, N + 1):
    for v in range(u + 1, N + 1):
        if abs(layer[u] - layer[v]) >= 2:
            removed[u][v] = True
            ans.append((u, v))

# Remove additional non-path edges arbitrarily until exactly M edges are removed.
for u in range(1, N + 1):
    if len(ans) >= M:
        break
    for v in range(u + 1, N + 1):
        if len(ans) >= M:
            break
        if removed[u][v]:
            continue
        if is_path_edge[u][v]:
            continue

        removed[u][v] = True
        ans.append((u, v))

print_yes()