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 nmax = (1 << 12) + 42, MX = 2.2e6;
int inp[nmax], where[nmax];
vector<int> tree[nmax * 4];
void build(int node, int l, int r) {
  for (int i = l; i <= r; i++) tree[node].push_back(where[i]);
  sort(tree[node].begin(), tree[node].end());
  if (l == r) return;
  int av = (l + r) / 2;
  build(node * 2, l, av);
  build(node * 2 + 1, av + 1, r);
}
pair<int, int> outp[MX];
int pointer;
map<pair<int, int>, int> mem[nmax * 4];
int query(int node, int lq, int rq) {
  int le = lower_bound(tree[node].begin(), tree[node].end(), lq) -
           tree[node].begin();
  int ri = upper_bound(tree[node].begin(), tree[node].end(), rq) -
           tree[node].begin() - 1;
  if (le > ri) return 0;
  if (le == ri) return tree[node][le];
  if (mem[node].count({le, ri})) return mem[node][{le, ri}];
  int mem_lhs = query(node * 2, lq, rq);
  int mem_rhs = query(node * 2 + 1, lq, rq);
  if (mem_lhs == 0) return mem_rhs;
  if (mem_rhs == 0) return mem_lhs;
  pointer++;
  outp[pointer] = {mem_lhs, mem_rhs};
  mem[node][{le, ri}] = pointer;
  return pointer;
}
int query_id[MX];
int main() {
  int n, q;
  scanf("%i%i", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%i", &inp[i]);
    where[inp[i]] = i;
  }
  build(1, 1, n);
  pointer = n;
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%i%i", &l, &r);
    query_id[i] = query(1, l, r);
  }
  printf("%i\n", pointer);
  for (int i = n + 1; i <= pointer; i++)
    printf("%i %i\n", outp[i].first, outp[i].second);
  for (int i = 1; i <= q; i++) printf("%i\n", query_id[i]);
  return 0;
}