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;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int n, Q;
  cin >> n >> Q;
  vector<int> a(n);
  for (int& x : a) {
    cin >> x;
    x--;
  }
  int cnt = n;
  vector<pair<int, int> > ops;
  auto merge = [&](int r, int s) -> int {
    if (r == 0) return s;
    if (s == 0) return r;
    cnt++;
    ops.push_back({r, s});
    int z = cnt;
    return z;
  };
  const int B = 8;
  const int T = 12;
  vector<vector<int> > need(1 << T);
  vector<map<pair<int, int>, int> > how(1 << T);
  for (int i = 0; i < n; i++) {
    need[a[i]] = {i};
    how[a[i]][{0, 0}] = i + 1;
  }
  for (int l = 0; l < B; l++) {
    vector<vector<int> > new_need(1 << (T - l - 1));
    vector<map<pair<int, int>, int> > new_how(1 << (T - l - 1));
    for (int j = 0; j < (1 << (T - l)); j += 2) {
      vector<int> r = need[j];
      r.insert(r.end(), need[j + 1].begin(), need[j + 1].end());
      need[j].push_back(-1);
      need[j + 1].push_back(-1);
      sort(r.begin(), r.end());
      new_need[j / 2] = r;
      vector<pair<int, int> > where(r.size());
      int i0 = 0, i1 = 0;
      for (int q = 0; q < (int)r.size(); q++) {
        if (need[j][i0] == r[q]) {
          where[q] = {0, i0};
          i0++;
        } else {
          where[q] = {1, i1};
          i1++;
        }
      }
      for (int st = 0; st < (int)r.size(); st++) {
        int first0 = -1, first1 = -1, last0 = -1, last1 = -1;
        for (int en = st; en < (int)r.size(); en++) {
          if (where[en].first == 0) {
            last0 = where[en].second;
            if (first0 == -1) first0 = where[en].second;
          } else {
            last1 = where[en].second;
            if (first1 == -1) first1 = where[en].second;
          }
          int s0 = 0, s1 = 0;
          if (first0 != -1) s0 = how[j][{first0, last0}];
          if (first1 != -1) s1 = how[j + 1][{first1, last1}];
          new_how[j / 2][{st, en}] = merge(s0, s1);
        }
      }
    }
    need = new_need;
    how = new_how;
  }
  vector<int> ans;
  for (int i = 0; i < Q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    int cur = 0;
    for (int j = 0; j < (1 << (T - B)); j++) {
      int s = lower_bound(need[j].begin(), need[j].end(), l) - need[j].begin();
      int e =
          lower_bound(need[j].begin(), need[j].end(), r + 1) - need[j].begin();
      if (s < e) {
        cur = merge(cur, how[j][{s, e - 1}]);
      }
    }
    ans.push_back(cur);
  }
  cout << cnt << '\n';
  for (pair<int, int> r : ops) {
    cout << r.first << ' ' << r.second << '\n';
  }
  for (int v : ans) cout << v << ' ';
  cout << '\n';
}