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

// n = 8, 9, 10 的方案

int A[8][6] = {
    0, 0, 9, 4, 0, 20,
    0, 20, 9, 4, 10, 20,
    10, 20, 9, 4, 11, 4,
    10, 20, 11, 4, 20, 20,
    20, 0, 20, 20, 11, 4,
    0, 0, 10, 0, 9, 4,
    10, 0, 11, 4, 9, 4,
    10, 0, 20, 0, 11, 4
};

int B[9][6] = {
    0, 0, 20, 0, 16, 12,
    0, 0, 16, 12, 0, 20,
    0, 20, 12, 14, 13, 20,
    20, 0, 20, 13, 16, 12,
    16, 16, 16, 12, 20, 13,
    16, 16, 20, 13, 20, 20,
    16, 16, 20, 20, 13, 20,
    16, 16, 13, 20, 12, 14,
    16, 16, 12, 14, 16, 12
};

int C[10][6] = {
    0, 0, 10, 8, 0, 20,
    20, 0, 10, 8, 20, 20,
    0, 20, 10, 8, 20, 20,
    0, 0, 8, 0, 5, 4,
    10, 3, 5, 4, 8, 0, 
    10, 3, 10, 8, 5, 4,
    10, 3, 8, 0, 12, 0,
    10, 3, 12, 0, 15, 4,
    10, 3, 15, 4, 10, 8,
    20, 0, 15, 4, 12, 0
};

void solve(int A[][6], int X, int Y) {
    // 用队列轮流把每个三角形拆成 4 个
    // 每个坐标后面都至少 7 个零，而所有三角形拆一轮数量乘 4，所以拆两轮就够了，无精度问题
    queue<vector<int>> q;
    for (int i = 0; i < X; i++) {
        vector<int> vec;
        for (int j = 0; j < 6; j++) vec.push_back(A[i][j] * ((int) 5e7));
        q.push(vec);
    }

    while (q.size() < Y) {
        vector<int> vec = q.front(); q.pop();
        int x1 = vec[0], y1 = vec[1];
        int x2 = vec[2], y2 = vec[3];
        int x3 = vec[4], y3 = vec[5];
        int x4 = (x1 + x2) / 2, y4 = (y1 + y2) / 2;
        int x5 = (x2 + x3) / 2, y5 = (y2 + y3) / 2;
        int x6 = (x3 + x1) / 2, y6 = (y3 + y1) / 2;
        q.push({x1, y1, x4, y4, x6, y6});
        q.push({x2, y2, x5, y5, x4, y4});
        q.push({x3, y3, x6, y6, x5, y5});
        q.push({x4, y4, x5, y5, x6, y6});
    }

    while (!q.empty()) {
        vector<int> vec = q.front(); q.pop();
        for (int i = 0; i < 6; i++) printf("%d%c", vec[i], "\n "[i < 5]);
    }
}

int main() {
    int n; scanf("%d", &n);
    if (n < 8) {
        printf("No\n");
        return 0;
    }

    printf("Yes\n");
    if (n % 3 == 2) solve(A, 8, n);
    else if (n % 3 == 0) solve(B, 9, n);
    else solve(C, 10, n);
    return 0;
}