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;
struct seg {
  vector<int> id;
  map<pair<int, int>, int> st;
  int getid(int l, int r) {
    int L = lower_bound(id.begin(), id.end(), l) - id.begin();
    int R = upper_bound(id.begin(), id.end(), r) - id.begin() - 1;
    if (L > R) return -1;
    return st[make_pair(L, R)];
  }
} tr[20000];
vector<pair<int, int> > res;
int cur;
int merge(int x, int y) {
  if (x == -1 || y == -1)
    return max(x, y);
  else {
    res.emplace_back(x, y);
    cur++;
    return cur;
  }
}
int blk, tot, Tid = 0;
int n, q, a[5000], Id[5000], l[5000], r[5000];
int solve(int l, int r, vector<int> id) {
  Tid++;
  int k;
  tr[k = Tid].id = id;
  if (l == r) {
    tr[Tid].st[pair<int, int>(0, 0)] = id[0];
    return Tid;
  }
  int mid = (l + r) >> 1;
  vector<int> lid, rid;
  for (auto x : id)
    if (a[x] <= mid)
      lid.push_back(x);
    else
      rid.push_back(x);
  int ls = solve(l, mid, lid), rs = solve(mid + 1, r, rid);
  for (int i = 0; i < id.size(); i++)
    for (int j = i; j < id.size(); j++)
      tr[k].st[pair<int, int>(i, j)] =
          merge(tr[ls].getid(id[i], id[j]), tr[rs].getid(id[i], id[j]));
  return k;
}
int main() {
  scanf("%d%d", &n, &q);
  blk = sqrt(q);
  cur = n;
  for (int i = 1; i <= (tot = n / blk); i++)
    l[i] = r[i - 1] + 1, r[i] = i * blk;
  if (tot * blk != n) l[++tot] = r[tot - 1] + 1, r[tot] = n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= tot; i++) {
    vector<int> id;
    for (int j = 1; j <= n; j++)
      if (l[i] <= a[j] && a[j] <= r[i]) id.push_back(j);
    Id[i] = solve(l[i], r[i], id);
  }
  vector<int> ans;
  for (int l, r; q--;) {
    scanf("%d%d", &l, &r);
    int rest = -1;
    for (int i = 1; i <= tot; i++) rest = merge(rest, tr[Id[i]].getid(l, r));
    ans.push_back(rest);
  }
  printf("%d\n", cur);
  for (auto x : res) printf("%d %d\n", x.first, x.second);
  for (int x : ans) printf("%d ", x);
}