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;
const int Q = (1 << 16);
const int K = 2200000;
map<array<int, 4>, int> id;
array<int, 2> par[K];
int value[N];
int ans[Q];
int m;
int get(int l, int r, int a, int b) {
  for (; l <= r && (value[l] < a || value[l] > b); ++l)
    ;
  for (; l <= r && (value[r] < a || value[r] > b); --r)
    ;
  if (l > r) {
    return 0;
  }
  if (id.count({l, r, a, b}) == 0) {
    if (a == b) {
      id[{l, r, a, b}] = l;
    } else {
      int c = (a + b) / 2;
      int u = get(l, r, a, c);
      int v = get(l, r, c + 1, b);
      if (u == 0 || v == 0) {
        id[{l, r, a, b}] = u ^ v;
      } else {
        id[{l, r, a, b}] = ++m;
        par[m] = {u, v};
      }
    }
  }
  return id[{l, r, a, b}];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  m = n;
  for (int i = 1; i <= n; ++i) {
    cin >> value[i];
  }
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    ans[i] = get(l, r, 1, n);
  }
  cout << m << "\n";
  for (int i = n + 1; i <= m; ++i) {
    for (auto j : par[i]) {
      cout << j << " ";
    }
    cout << "\n";
  }
  for (int i = 0; i < q; ++i) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}