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;
vector<int> inv;
int cnt;
vector<int> u, v;
struct Tree {
  int lo, hi;
  Tree *bot, *top;
  vector<int> numdwn;
  unordered_map<long long, int> mp;
  Tree(int *l, int *r, int x, int y) {
    lo = x, hi = y;
    if (lo == hi || l >= r) return;
    int mid = (lo + hi) / 2, cntdwn = 0;
    numdwn.reserve(r - l);
    for (int *p = l; p != r; ++p) {
      if (*p <= mid) ++cntdwn;
      numdwn.push_back(cntdwn);
    }
    int *divider = stable_partition(l, r, [mid](int v) { return v <= mid; });
    bot = new Tree(l, divider, lo, mid);
    top = new Tree(divider, r, mid + 1, hi);
  }
  int idof(int l, int r) {
    if (l > r) return -1;
    if (lo == hi) return inv[lo];
    long long code = ((long long)l) * n + r;
    if (mp.count(code)) return mp[code];
    int numbl = 0;
    if (l > 0) numbl = numdwn[l - 1];
    int numbr = numdwn[r], numb = numbr - numbl;
    int botid = bot->idof(numbl, numbr - 1);
    int topid = top->idof(l - numbl, r - numbr);
    int ans = -1;
    if (botid == -1)
      ans = topid;
    else if (topid == -1)
      ans = botid;
    else {
      ans = cnt++;
      u.push_back(botid);
      v.push_back(topid);
    }
    mp[code] = ans;
    return ans;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < (n); ++i) {
    cin >> a[i];
    --a[i];
  }
  inv.resize(n);
  for (int i = 0; i < (n); ++i) inv[a[i]] = i;
  auto acopy = a;
  cnt = n;
  Tree tree(&acopy[0], &acopy[0] + n, 0, n - 1);
  vector<int> lq(q), rq(q);
  for (int i = 0; i < (q); ++i) {
    cin >> lq[i] >> rq[i];
    --lq[i];
    --rq[i];
  }
  vector<int> ks(q);
  for (int i = 0; i < (q); ++i) {
    ks[i] = tree.idof(lq[i], rq[i]);
  }
  cout << cnt << "\n";
  for (int i = 0; i < ((int)(u).size()); ++i) {
    cout << u[i] + 1 << " " << v[i] + 1 << "\n";
  }
  for (int i = 0; i < (q); ++i) cout << ks[i] + 1 << " ";
  cout << "\n";
}