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 CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
template <class T, class U>
inline void checkmin(T &x, U y) {
  if (y < x) x = y;
}
template <class T, class U>
inline void checkmax(T &x, U y) {
  if (y > x) x = y;
}
template <class T, class U>
inline bool ifmax(T &x, U y) {
  if (y > x) return x = y, true;
  return false;
}
template <class T, class U>
inline bool ifmin(T &x, U y) {
  if (y < x) return x = y, true;
  return false;
}
long long fstpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
long long powm(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
void fukof() {}
void problem() {}
const long long mod = 1e9 + 7;
const long long N = 100005;
const long long inf = 1e9;
void solve() {
  long long n;
  cin >> n;
  long long a[n + 1], sum = 0;
  for (long long i = 0; i < (n); i++) cin >> a[i + 1], sum += a[i + 1];
  if (sum < 2 * n - 2) {
    cout << "NO\n";
    return;
  }
  cout << "YES ";
  long long cnt2 = 0, cnt1 = 0;
  vector<long long> a1, a2;
  for (long long i = 0; i < (n); i++) {
    if (a[i + 1] > 1)
      cnt2++, a2.push_back(i + 1);
    else
      cnt1++, a1.push_back(i + 1);
  }
  cout << cnt2 + (cnt1 >= 2 ? 2 : cnt1) - 1 << endl;
  cout << n - 1 << endl;
  if (cnt2 and cnt1) {
    a[a2[0]]--;
    a[a1[0]]--;
    cnt1--;
    cout << a2[0] << " " << a1[0] << endl;
    if (cnt1) {
      a[a2.size() - 1]--;
      a[a1[1]]--;
      cnt1--;
      cout << a2[a2.size() - 1] << " " << a1[1] << endl;
    }
  }
  for (long long i = 0; i < (a2.size() - 1); i++) {
    a[a2[i]]--;
    a[a2[i + 1]]--;
    cout << a2[i] << " " << a2[i + 1] << endl;
  }
  long long ptr = 0, ptr2 = a1.size() - 1;
  while (cnt1 > 0) {
    if (a[a2[ptr]] > 0) {
      a[a2[ptr]]--;
      a[a1[ptr2]]--;
      cout << a2[ptr] << " " << a1[ptr2] << endl;
      ptr2--;
      cnt1--;
    } else
      ptr++;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int tc = 1;
  while (tc--) {
    problem();
    fukof();
    solve();
  }
  return 0;
}