Problem: Graph constructive problems are back! This time the graph you are asked to build should match the following properties.

The graph is connected if and only if there exists a path between every pair of vertices.

The diameter (aka "longest shortest path") of a connected undirected graph is the maximum number of edges in the shortest path between any pair of its vertices.

The degree of a vertex is the number of edges incident to it.

Given a sequence of n integers a_1, a_2, ..., a_n construct a connected undirected graph of n vertices such that:

  * the graph contains no self-loops and no multiple edges; 
  * the degree d_i of the i-th vertex doesn't exceed a_i (i.e. d_i ≤ a_i); 
  * the diameter of the graph is maximum possible. 



Output the resulting graph or report that no solution exists.

Input

The first line contains a single integer n (3 ≤ n ≤ 500) — the number of vertices in the graph.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n - 1) — the upper limits to vertex degrees.

Output

Print "NO" if no graph can be constructed under the given conditions.

Otherwise print "YES" and the diameter of the resulting graph in the first line.

The second line should contain a single integer m — the number of edges in the resulting graph.

The i-th of the next m lines should contain two integers v_i, u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the description of the i-th edge. The graph should contain no multiple edges — for each pair (x, y) you output, you should output no more pairs (x, y) or (y, x).

Examples

Input


3
2 2 2


Output


YES 2
2
1 2
2 3


Input


5
1 4 1 1 1


Output


YES 2
4
1 2
3 2
4 2
5 2


Input


3
1 1 1


Output


NO

Note

Here are the graphs for the first two example cases. Both have diameter of 2.

<image> d_1 = 1 ≤ a_1 = 2

d_2 = 2 ≤ a_2 = 2

d_3 = 1 ≤ a_3 = 2 

<image> d_1 = 1 ≤ a_1 = 1

d_2 = 4 ≤ a_2 = 4

d_3 = 1 ≤ a_3 = 1

d_4 = 1 ≤ a_4 = 1 

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Vector;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Stack;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Vadim
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        D solver = new D();
        solver.solve(1, in, out);
        out.close();
    }

    static class D {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.ni();
            int[] a = in.na(n);
            Stack<Integer> leaves = new Stack<Integer>();
            for (int i = 0; i < n; i++) {
                if (a[i] == 1)
                    leaves.push(i);
            }
            if (leaves.size() < 2) {
                out.println("YES " + (n - 1));
                out.println(n - 1);
                int first = 0;
                if (leaves.size() == 1) {
                    first = leaves.pop();
                }
                int prev = first;
                for (int i = 0; i < n; i++) {
                    if (i != first) {
                        out.println((prev + 1) + " " + (i + 1));
                        prev = i;
                    }
                }
            } else {
                int[] g = new int[n];
                int first = leaves.pop();
                int last = leaves.pop();
                int prev = first;
                int d = 1;
                for (int i = 0; i < n; i++) {
                    if (i != first && i != last && a[i] != 1) {
                        g[i] = prev;
                        d++;
                        if (a[i] > 2 && !leaves.isEmpty()) {
                            for (int j = 0; j < a[i] - 2 && !leaves.isEmpty(); j++) {
                                int l = leaves.pop();
                                g[l] = i;
                            }
                        }
                        prev = i;
                    }
                }
                g[last] = prev;
                if (!leaves.isEmpty()) {
                    out.println("NO");
                } else {
                    out.println("YES " + d);
                    out.println(n - 1);
                    for (int i = 0; i < n; i++) {
                        if (i != first) {
                            out.println((i + 1) + " " + (g[i] + 1));
                        }
                    }
                }
            }

        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String ns() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int ni() {
            return Integer.parseInt(ns());
        }

        public int[] na(int n) {
            int[] ret = new int[n];
            for (int i = 0; i < n; i++) {
                ret[i] = ni();
            }
            return ret;
        }

    }
}