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;
class TaskH {
 public:
  int cnt;
  vector<pair<int, int> > ops;
  vector<int> innerSolve(const vector<int>& a, const vector<pair<int, int> >& b,
                         int low, int high) {
    vector<int> res(b.size(), -1);
    map<pair<int, int>, int> encode;
    vector<pair<int, int> > nb;
    for (int i = 0; i < b.size(); ++i) {
      auto p = b[i];
      while (p.first <= p.second && (a[p.first] < low || a[p.first] > high))
        ++p.first;
      while (p.first <= p.second && (a[p.second] < low || a[p.second] > high))
        --p.second;
      if (p.first > p.second) continue;
      if (encode.find(p) != encode.end()) {
        int idx = encode[p];
        res[i] = idx;
      } else {
        int idx = encode.size();
        nb.push_back(p);
        encode[p] = idx;
        res[i] = idx;
      }
    }
    if (!nb.empty()) {
      vector<int> nres(nb.size());
      if (low == high) {
        assert(nb.size() == 1);
        nres[0] = nb[0].first + 1;
      } else {
        int mid = (low + high) / 2;
        vector<int> r1 = innerSolve(a, nb, low, mid);
        vector<int> r2 = innerSolve(a, nb, mid + 1, high);
        for (int i = 0; i < nres.size(); ++i) {
          if (r1[i] == 0)
            nres[i] = r2[i];
          else if (r2[i] == 0)
            nres[i] = r1[i];
          else {
            ops.emplace_back(r1[i], r2[i]);
            nres[i] = a.size() + ops.size();
          }
        }
      }
      for (int i = 0; i < res.size(); ++i)
        if (res[i] >= 0) res[i] = nres[res[i]];
    }
    for (int i = 0; i < res.size(); ++i)
      if (res[i] < 0) res[i] = 0;
    return res;
  }
  void solveOne(istream& in, ostream& out) {
    int n, q;
    in >> n >> q;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      in >> a[i];
    }
    vector<pair<int, int> > b(q);
    for (int i = 0; i < q; ++i) {
      in >> b[i].first >> b[i].second;
      --b[i].first;
      --b[i].second;
    }
    cnt = n;
    ops.clear();
    vector<int> res = innerSolve(a, b, 1, n);
    out << a.size() + ops.size() << "\n";
    for (auto x : ops) {
      out << x.first << " " << x.second << "\n";
    }
    for (int x : res) {
      out << x << " ";
    }
    out << "\n";
  }
  void solve(std::istream& in, std::ostream& out) {
    int nt = 1;
    for (int it = 0; it < nt; ++it) {
      solveOne(in, out);
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  TaskH solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}