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 n;
const int N = (1 << 12) + 5, K = (1 << 8) + 5;
int tab[N], tab2[N][K], tab3[N];
vector<pair<int, int> > state;
vector<int> V[K];
vector<vector<int> > dp[K];
vector<vector<int> > gen(vector<int> vec, int l, int r) {
  if (l == r) {
    vector<vector<int> > res(1);
    res[0].push_back(tab3[l]);
    return res;
  }
  int m = (l + r) / 2;
  vector<int> V1, V2;
  for (int i : vec) {
    if (i <= m)
      V1.push_back(i);
    else
      V2.push_back(i);
  }
  vector<vector<int> > res1 = gen(V1, l, m), res2 = gen(V2, m + 1, r), res(n);
  int p1 = 0, p2 = 0, k1 = 0, k2 = 0;
  for (int i = 0; i < vec.size(); i++) {
    for (int j = i; j < vec.size(); j++) {
      if (vec[j] <= m)
        k1++;
      else
        k2++;
      if (p1 == k1) {
        res[i].push_back(res2[p2][k2 - p2 - 1]);
      } else {
        if (p2 == k2) {
          res[i].push_back(res1[p1][k1 - p1 - 1]);
        } else {
          state.push_back({res1[p1][k1 - p1 - 1], res2[p2][k2 - p2 - 1]});
          res[i].push_back(state.size() + n);
        }
      }
    }
    if (vec[i] <= m)
      p1++;
    else
      p2++;
    k1 = p1;
    k2 = p2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> n >> q;
  int k = (1 << 8), t = (n + k - 1) / k;
  for (int i = 0; i < n; i++) {
    cin >> tab[i];
    tab[i]--;
    tab3[tab[i]] = i + 1;
    V[tab[i] / k].push_back(tab[i]);
    for (int j = 0; j < t; j++) {
      tab2[i + 1][j] = tab2[i][j];
    }
    tab2[i + 1][tab[i] / k]++;
  }
  for (int i = 0; i < t; i++) {
    dp[i] = gen(V[i], i * k, min(n - 1, k * (i + 1) - 1));
  }
  vector<int> ans;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    int act = -1;
    for (int i = 0; i < t; i++) {
      if (tab2[l][i] != tab2[r][i]) {
        if (act == -1)
          act = dp[i][tab2[l][i]][tab2[r][i] - tab2[l][i] - 1];
        else {
          state.push_back(
              {act, dp[i][tab2[l][i]][tab2[r][i] - tab2[l][i] - 1]});
          act = state.size() + n;
        }
      }
    }
    ans.push_back(act);
  }
  cout << state.size() + n << "\n";
  for (auto i : state) {
    cout << i.first << " " << i.second << "\n";
  }
  for (int i : ans) {
    cout << i << " ";
  }
}