Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 12;
const int max_size = 1 << 8;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    --a[i];
  }
  vector<int> pos(n);
  for (int i = 0; i < n; ++i) {
    pos[a[i]] = i;
  }
  int cnt = n;
  vector<pair<int, int>> ans;
  auto get = [&](int f, int s) {
    if (f == 0) return s;
    if (s == 0) return f;
    ans.push_back({f, s});
    return ++cnt;
  };
  vector<map<pair<int, int>, int>> id;
  function<void(int, int, int)> build = [&](int l, int r, int v) {
    if (l < r) {
      int m = (l + r) >> 1;
      build(l, m, 2 * v + 1);
      build(m + 1, r, 2 * v + 2);
      if (r - l + 1 <= max_size) {
        vector<int> where;
        for (int i = l; i <= r; ++i) {
          where.push_back(pos[i]);
        }
        sort(where.begin(), where.end());
        for (int L = 0; L < (int)where.size(); ++L) {
          int l_f = n - 1, r_f = 0, l_s = n - 1, r_s = 0;
          for (int R = L; R < (int)where.size(); ++R) {
            if (a[where[R]] <= m) {
              l_f = min(l_f, where[R]);
              r_f = max(r_f, where[R]);
            } else {
              l_s = min(l_s, where[R]);
              r_s = max(r_s, where[R]);
            }
            id[v][{where[L], where[R]}] =
                get(id[2 * v + 1][{l_f, r_f}], id[2 * v + 2][{l_s, r_s}]);
          }
        }
      }
    } else {
      while (id.size() <= v) id.push_back({});
      id[v][{pos[l], pos[l]}] = pos[l] + 1;
    }
  };
  build(0, n - 1, 0);
  function<int(int, int, int, int, int)> answer = [&](int ql, int qr, int l,
                                                      int r, int v) {
    if (r - l + 1 > max_size) {
      int m = (l + r) >> 1;
      int f = answer(ql, qr, l, m, 2 * v + 1);
      int s = answer(ql, qr, m + 1, r, 2 * v + 2);
      return get(f, s);
    } else {
      int l_f = n - 1, r_f = 0;
      for (int i = l; i <= r; ++i) {
        int x = pos[i];
        if (ql <= x && x <= qr) {
          l_f = min(l_f, x);
          r_f = max(r_f, x);
        }
      }
      return id[v][{l_f, r_f}];
    }
  };
  vector<int> who;
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    who.push_back(answer(l, r, 0, n - 1, 0));
  }
  cout << cnt << '\n';
  for (auto p : ans) {
    cout << p.first << ' ' << p.second << '\n';
  }
  for (int x : who) cout << x << ' ';
}