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) + 1, B = (1 << 4), SZ = (1 << 8);
const int M = 2200050;
int cntn, n, q;
pair<int, int> p[M];
vector<int> sts[M];
int a[N];
vector<vector<int> > idxs[B];
vector<int> xs[B];
long long res[(1 << 16) + 5];
int add_set(int u, int v) {
  if (u == -1 || v == -1) return (u == -1 ? v : u);
  p[cntn] = {u, v};
  return cntn++;
}
vector<vector<int> > solve(vector<int>& ids) {
  if (ids.empty()) return {};
  if ((int)(ids).size() == 1) {
    return {{ids[0]}};
  }
  vector<vector<int> > res((int)(ids).size(),
                           vector<int>((int)(ids).size(), 0));
  vector<int> ys;
  for (int i : ids) ys.push_back(a[i]);
  sort(ys.begin(), ys.end());
  int mid = ys[((int)(ys).size() - 1) / 2];
  vector<int> Lid, Rid;
  for (int i : ids) {
    if (a[i] <= mid)
      Lid.push_back(i);
    else
      Rid.push_back(i);
  }
  vector<vector<int> > L = solve(Lid), R = solve(Rid);
  int la = 0, ra = 0;
  for (int i = 0; i < (int)(ids).size(); i++) {
    while (la < (int)(Lid).size() && Lid[la] < ids[i]) la++;
    while (ra < (int)(Rid).size() && Rid[ra] < ids[i]) ra++;
    int lb = 0, rb = 0;
    for (int j = i; j < (int)(ids).size(); j++) {
      while (lb < (int)(Lid).size() && Lid[lb] <= ids[j]) lb++;
      while (rb < (int)(Rid).size() && Rid[rb] <= ids[j]) rb++;
      if (lb > la && rb > ra) {
        res[i][j] = add_set(L[la][lb - 1], R[ra][rb - 1]);
      } else if (lb == la) {
        assert(rb != ra);
        res[i][j] = R[ra][rb - 1];
      } else {
        res[i][j] = L[la][lb - 1];
      }
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> q;
  cntn = n;
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    a[i]--;
    xs[a[i] / SZ].push_back(i);
  }
  for (int i = 0; i < (n); ++i) sts[i].push_back(i);
  for (int i = 0; i < (B); ++i) {
    idxs[i] = solve(xs[i]);
  }
  for (int i = 0; i < (q); ++i) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    int curset = -1;
    for (int b = 0; b < (B); ++b) {
      int lb =
          (int)(lower_bound(xs[b].begin(), xs[b].end(), l) - xs[b].begin());
      int rb =
          (int)(upper_bound(xs[b].begin(), xs[b].end(), r) - xs[b].begin());
      if (rb > lb) {
        curset = add_set(curset, idxs[b][lb][rb - 1]);
      }
    }
    res[i] = curset + 1;
  }
  cout << cntn << endl;
  for (int i = n; i < (cntn); ++i) {
    cout << p[i].first + 1 << " " << p[i].second + 1 << "\n";
  }
  for (int i = 0; i < (q); ++i) cout << res[i] << " ";
  cout << '\n';
}