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, m;
int cnt;
const int N = (1 << 12) + 20;
vector<vector<int> > Block[17];
vector<pair<int, int> > operators;
inline int merge(int a, int b) {
  if (!a || !b) return a | b;
  operators.push_back(make_pair(a, b));
  return ++cnt;
}
int Limit;
int num;
int bel[N];
int a[N];
inline vector<vector<int> > get_idx(vector<int> idx, int all_rngs) {
  int n = idx.size() - 1;
  if (all_rngs == Limit) {
    ++num;
    for (int i = 1; i <= n; i++) bel[idx[i]] = num;
  }
  if (all_rngs < Limit) {
    vector<int> b;
    for (int i = 1; i <= n; i++) b.push_back(a[idx[i]]);
    sort(b.begin(), b.end());
    int Split_Number = b[n / 2 - 1];
    vector<int> Lft(1, 0), Rgt(1, 0);
    for (int i = 1; i <= n; i++)
      if (a[idx[i]] <= Split_Number)
        Lft.push_back(idx[i]);
      else
        Rgt.push_back(idx[i]);
    vector<vector<int> > L = get_idx(Lft, all_rngs * 2);
    vector<vector<int> > R = get_idx(Rgt, all_rngs * 2);
    return vector<vector<int> >();
  }
  vector<vector<int> > ret(n + 2, vector<int>(n + 2));
  if (n == 1) {
    ret[1][1] = idx[1];
  } else {
    vector<int> b;
    for (int i = 1; i <= n; i++) b.push_back(a[idx[i]]);
    sort(b.begin(), b.end());
    int Split_Number = b[n / 2 - 1];
    vector<int> Lft(1, 0), Rgt(1, 0);
    for (int i = 1; i <= n; i++)
      if (a[idx[i]] <= Split_Number)
        Lft.push_back(idx[i]);
      else
        Rgt.push_back(idx[i]);
    vector<vector<int> > L = get_idx(Lft, all_rngs * 2);
    vector<vector<int> > R = get_idx(Rgt, all_rngs * 2);
    vector<int> pref[2];
    pref[0] = pref[1] = vector<int>(n + 1, 0);
    for (int i = 1; i <= n; i++) {
      pref[0][i] = pref[0][i - 1], pref[1][i] = pref[1][i - 1];
      if (a[idx[i]] <= Split_Number)
        pref[0][i]++;
      else
        pref[1][i]++;
    }
    for (int i = 1; i <= n; i++)
      for (int j = i; j <= n; j++) {
        if (i == j)
          ret[i][j] = idx[i];
        else {
          ret[i][j] = merge(L[pref[0][i - 1] + 1][pref[0][j]],
                            R[pref[1][i - 1] + 1][pref[1][j]]);
        }
      }
  }
  if (all_rngs == Limit) {
    Block[num] = ret;
  }
  return ret;
}
int pref[17][N];
int main() {
  cin >> n >> m;
  cnt = n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  Limit = 1;
  while (Limit * 2 <= n && Limit < 16) Limit <<= 1;
  vector<int> q(1, 0);
  for (int i = 1; i <= n; i++) q.push_back(i);
  get_idx(q, 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= num; j++) pref[j][i] = pref[j][i - 1];
    pref[bel[i]][i]++;
  }
  vector<int> ans;
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    int cur = 0;
    for (int i = 1; i <= num; i++) {
      int _l = pref[i][l - 1] + 1, _r = pref[i][r];
      cur = merge(cur, Block[i][_l][_r]);
    }
    ans.push_back(cur);
  }
  cout << cnt << endl;
  for (size_t i = 0; i < operators.size(); i++) {
    printf("%d %d\n", operators[i].first, operators[i].second);
  }
  for (size_t i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}