Problem: An undirected graph is called k-regular, if the degrees of all its vertices are equal k. An edge of a connected graph is called a bridge, if after removing it the graph is being split into two connected components.

Build a connected undirected k-regular graph containing at least one bridge, or else state that such graph doesn't exist.

Input

The single line of the input contains integer k (1 ≤ k ≤ 100) — the required degree of the vertices of the regular graph.

Output

Print "NO" (without quotes), if such graph doesn't exist. 

Otherwise, print "YES" in the first line and the description of any suitable graph in the next lines.

The description of the made graph must start with numbers n and m — the number of vertices and edges respectively. 

Each of the next m lines must contain two integers, a and b (1 ≤ a, b ≤ n, a ≠ b), that mean that there is an edge connecting the vertices a and b. A graph shouldn't contain multiple edges and edges that lead from a vertex to itself. A graph must be connected, the degrees of all vertices of the graph must be equal k. At least one edge of the graph must be a bridge. You can print the edges of the graph in any order. You can print the ends of each edge in any order.

The constructed graph must contain at most 106 vertices and 106 edges (it is guaranteed that if at least one graph that meets the requirements exists, then there also exists the graph with at most 106 vertices and at most 106 edges). 

Examples

Input

1


Output

YES
2 1
1 2

Note

In the sample from the statement there is a suitable graph consisting of two vertices, connected by a single edge.

Solution (in CPP):

Solution:
// author: Kaidul
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(__typeof(n) i = 0; i < (n); i++)
#define rrep(i, n) for(__typeof(n) i = (n) - 1; i >= 0; --i)
#define rep1(i, n) for(__typeof(n) i = 1; i <= (n); i++)
#define FOR(i, a, b) for(__typeof(b) i = (a); i <= (b); i++)
#define forstl(i, s) for (__typeof ((s).end ()) i = (s).begin (); i != (s).end (); ++i)
#define INF (1 << 30)
#define PI acos(-1.0)
#define pb push_back
#define ppb pop_back
#define all(x) x.begin(), x.end()
#define mem(x, y) memset(x, y, sizeof x)
#define eps 1e-9
#define pii pair<int, int>
#define couple make_pair
#define X first
#define Y second
#define vi vector<int>
#define vpii vector< pii >
#define si set<int>
#define SDi(x) sf("%d", &x)
#define SD2(x, y) sf("%d %d", &x, &y)
#define SD3(x, y, z) sf("%d %d %d", &x, &y, &z)
#define SDs(x) sf("%s", x)
#define pf printf
#define print(x) pf("%d ", x)
#define println(x) pf("%d\n", x)
#define output(x, y); pf("Case %d: %d", ++x, y)
#define newLine pf("\n")
#define sf scanf
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
#if ( _WIN32 or __WIN32__ )
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
#define SDl(x) sf(LLD, &x)
#define MAX5 100000
#define MAX7 10000000
#define MAX9 1000000000
#define MOD7 (MAX7 + 7)
#define MOD9 (MAX9 + 9)
typedef long long i64;
typedef unsigned long long ui64;
const i64 INF64 = (i64)1E18;

class TimeTracker {
    clock_t start, end;
public:
    TimeTracker() {
        start = clock();
    }
    ~TimeTracker() {
        end = clock();
        fprintf(stderr, "%.3lf s\n", (double)(end - start) / CLOCKS_PER_SEC);
    }
};

string toString(int num) {
    ostringstream convert;
    convert << num;
    return convert.str();
}
struct Point {
    int x, y;
    Point(): x(0), y(0) {}
    Point(int a, int b): x(a), y(b) {}
    bool operator < (const Point& other) const {
        return x < other.x;
    }
};
// BitMask
int Set(int N, int pos) {
    return N = N | (1 << pos);
}
int Reset(int N, int pos) {
    return N = N & ~(1 << pos);
}
int Check(int N, int pos) {
    return (N & (1 << pos));
}
int toggle(int N, int pos) {
    if( Check(N, pos) )
        return N = Reset(N, pos);
    return N = Set(N, pos);
}

// direction array
//int dx[] = {0, -1, 0, 1};
//int dy[] = {-1, 0, 1, 0};
//int Dx[] = {0, -1, -1, -1, 0, 1, 1, 1};
//int Dy[] = {-1, -1, 0, 1, 1, 1, 0, -1};
//int row, col;
//bool isValid(int i, int j) {
//    return i >= 0 and j >= 0 and i < row and j < col;
//}

/** Implementation **/
#define MAX 300
pair<int, int> degrees[MAX];

int main(void) {
#ifndef ONLINE_JUDGE
//     READ("input.txt");
//     WRITE("output.txt");
#endif
    /*
    int tcase, caseNo = 0;
    int k;
    while(SDi(k) == 1) {
        if(k >= 2) {
            puts("NO");
        } else {
            puts("YES");
            printf("2 1\n");
            printf("1 2\n");
        }
    }
    **/
    int k;
    SDi(k);
    if(!(k & 1)) {
        puts("NO");
        return 0;
    }
    if(k == 1) {
        puts("YES");
        printf("2 1\n");
        printf("1 2\n");
        return 0;
    }
    int n = k + 2;
    int m = n * k;
    puts("YES");
    printf("%d %d\n", 2 * n, m);
    FOR(i, 1, n) {
        FOR(j, i + 1, n) {
            if(i == 1 and j > k) {
                continue;
            }
            if(j < k + 1 and j == i + 1 and (j & 1)) {
                continue;
            }
            printf("%d %d\n", i, j);
            printf("%d %d\n", n + i, n + j);
        }
    }
    printf("1 %d\n", n + 1);
    return EXIT_SUCCESS;
}