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 PYTHON):

Solution:
#!/usr/bin/env python
"""
This file is part of https://github.com/Cheran-Senthil/PyRival.

Copyright 2018 Cheran Senthilkumar all rights reserved,
Cheran Senthilkumar <hello@cheran.io>
Permission to use, modify, and distribute this software is given under the
terms of the MIT License.

"""
from __future__ import division, print_function

import cmath
import itertools
import math
import operator as op
# import random
import sys
from atexit import register
from bisect import bisect_left, bisect_right
# from collections import Counter, MutableSequence, defaultdict, deque
# from copy import deepcopy
# from decimal import Decimal
# from difflib import SequenceMatcher
# from fractions import Fraction
# from heapq import heappop, heappush

if sys.version_info[0] < 3:
    # from cPickle import dumps
    from io import BytesIO as stream
    # from Queue import PriorityQueue, Queue
else:
    # from functools import reduce
    from io import StringIO as stream
    from math import gcd
    # from pickle import dumps
    # from queue import PriorityQueue, Queue


if sys.version_info[0] < 3:
    class dict(dict):
        """dict() -> new empty dictionary"""
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)

        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)

        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)

    def gcd(x, y):
        """gcd(x, y) -> int
        greatest common divisor of x and y
        """
        while y:
            x, y = y, x % y
        return x

    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip


def sync_with_stdio(sync=True):
    """Set whether the standard Python streams are allowed to buffer their I/O.

    Args:
        sync (bool, optional): The new synchronization setting.

    """
    global input, flush

    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        input = lambda: sys.stdin.readline().rstrip('\r\n')

        sys.stdout = stream()
        register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))


def main():
    n = int(input())
    a = map(int, input().split(' '))

    degree_one_nodes = []
    other_nodes = []

    for i, ai in enumerate(a):
        if ai == 1:
            degree_one_nodes.append(i + 1)
        else:
            other_nodes.append((i + 1, ai))

    if sum(i[1] - 2 for i in other_nodes) + 2 < len(degree_one_nodes):
        print('NO')
        return

    print('YES', len(other_nodes) + min(2, len(degree_one_nodes)) - 1)

    edges = []

    for i in range(len(other_nodes) - 1):
        edges.append((other_nodes[i][0], other_nodes[i + 1][0]))

    try:
        edges.append((other_nodes[0][0], degree_one_nodes.pop()))
    except IndexError:
        pass

    try:
        edges.append((other_nodes[-1][0], degree_one_nodes.pop()))
    except IndexError:
        pass

    for i in range(len(other_nodes)):
        for _ in range(other_nodes[i][1] - 2):
            try:
                edges.append((other_nodes[i][0], degree_one_nodes.pop()))
            except IndexError:
                break

    print(len(edges))
    for edge in edges:
        print(*edge)

    return


if __name__ == '__main__':
    sync_with_stdio(False)
    main()