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;
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
  return os << "(" << x.first << "," << x.second << ")";
}
int n;
vector<int> pos;
vector<vector<int>> ids;
vector<map<pair<int, int>, int>> memo;
vector<pair<int, int>> ans;
void build(int sn, int s, int e) {
  if (s == e) {
    ids[sn] = {pos[s]};
    return;
  }
  int m = (s + e) >> 1, lsn = sn * 2, rsn = sn * 2 + 1;
  build(lsn, s, m);
  build(rsn, m + 1, e);
  ids[sn].resize(ids[lsn].size() + ids[rsn].size());
  merge(ids[lsn].begin(), ids[lsn].end(), ids[rsn].begin(), ids[rsn].end(),
        ids[sn].begin());
}
int query(int sn, int s, int e, int qs, int qe) {
  int l = lower_bound(ids[sn].begin(), ids[sn].end(), qs) - ids[sn].begin();
  int r = upper_bound(ids[sn].begin(), ids[sn].end(), qe) - ids[sn].begin() - 1;
  if (l > r) return -1;
  pair<int, int> v = {l, r};
  auto& mp = memo[sn];
  if (mp.count(v)) return mp[v];
  if (s == e) return mp[v] = ids[sn][0];
  int m = (s + e) >> 1, lsn = sn * 2, rsn = sn * 2 + 1;
  int x = query(lsn, s, m, qs, qe);
  int y = query(rsn, m + 1, e, qs, qe);
  if (x == -1) return mp[v] = y;
  if (y == -1) return mp[v] = x;
  ans.emplace_back(x, y);
  return mp[v] = ans.size() + n;
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  pos.resize(n + 1);
  int maxs = n == 1 ? 2 : 1 << (33 - __builtin_clz(n - 1));
  ids.resize(maxs);
  memo.resize(maxs);
  for (int i = 0; i < n; ++i) {
    int a;
    scanf("%d", &a);
    pos[a] = i + 1;
  }
  build(1, 1, n);
  vector<int> qans(q);
  for (int i = 0; i < q; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    qans[i] = query(1, 1, n, l, r);
  }
  printf("%d\n", ans.size() + n);
  for (auto& a : ans) printf("%d %d\n", a.first, a.second);
  for (int i = 0; i < q; ++i) printf("%d ", qans[i]);
}