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 MAXN = 4200;
const int MAXP = 2 * MAXN + 100;
const int MAXQ = (1 << 16) + 100;
const int MAXA = 2.5e6;
int N, Q;
int arr[MAXN];
int rloc[MAXN];
pair<int, int> rng[MAXQ];
int P, nd;
int rlo[MAXP], rhi[MAXP];
int nv[MAXN][MAXN], pv[MAXN][MAXN];
map<pair<int, int>, int> mm[MAXP];
int A;
int p1[MAXA], p2[MAXA];
int aloc[MAXQ];
int gsolve(int pp, int lo, int hi);
int qsolve(int pp, int lo, int hi) {
  int r1 = pp * 2, r2 = pp * 2 + 1;
  int g1 = gsolve(r1, lo, hi), g2 = gsolve(r2, lo, hi);
  if (g1 == -1) return g2;
  if (g2 == -1) return g1;
  p1[A] = g1;
  p2[A] = g2;
  A++;
  return A - 1;
}
int gsolve(int pp, int lo, int hi) {
  if (pp >= P) {
    if (pp - P >= N) return -1;
    int a = rloc[pp - P];
    if (a >= lo && a <= hi)
      return a;
    else
      return -1;
  }
  if (nv[pp][lo] > hi) return -1;
  lo = nv[pp][lo];
  hi = pv[pp][hi];
  if (mm[pp].find(make_pair(lo, hi)) != mm[pp].end())
    return mm[pp][make_pair(lo, hi)];
  int res = qsolve(pp, lo, hi);
  mm[pp][make_pair(lo, hi)] = res;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> Q;
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
    arr[i]--;
    rloc[arr[i]] = i;
  }
  for (int i = 0; i < Q; i++) {
    cin >> rng[i].first >> rng[i].second;
    rng[i].first--;
    rng[i].second--;
  }
  P = 1;
  nd = 0;
  while (P < N) {
    P *= 2;
    nd++;
  }
  rlo[1] = 0;
  rhi[1] = P - 1;
  for (int i = 2; i < 2 * P; i++) {
    int md = (rlo[i / 2] + rhi[i / 2]) / 2;
    if (i % 2 == 0) {
      rlo[i] = rlo[i / 2];
      rhi[i] = md;
    } else {
      rlo[i] = md + 1;
      rhi[i] = rhi[i / 2];
    }
  }
  for (int i = 1; i < P; i++) {
    nv[i][N] = N;
    for (int j = N - 1; j >= 0; j--) {
      if (arr[j] >= rlo[i] && arr[j] <= rhi[i])
        nv[i][j] = j;
      else
        nv[i][j] = nv[i][j + 1];
    }
    if (arr[0] >= rlo[i] && arr[0] <= rhi[i])
      pv[i][0] = 0;
    else
      pv[i][0] = -1;
    for (int j = 1; j < N; j++) {
      if (arr[j] >= rlo[i] && arr[j] <= rhi[i])
        pv[i][j] = j;
      else
        pv[i][j] = pv[i][j - 1];
    }
  }
  for (int i = 0; i < N; i++) {
    mm[arr[i] + P][make_pair(i, i)] = i;
  }
  A = N;
  for (int i = 0; i < Q; i++) {
    aloc[i] = gsolve(1, rng[i].first, rng[i].second);
  }
  cout << A << "\n";
  for (int i = N; i < A; i++) {
    cout << p1[i] + 1 << " " << p2[i] + 1 << "\n";
  }
  for (int i = 0; i < Q; i++) {
    if (i) cout << " ";
    cout << aloc[i] + 1;
  }
  cout << "\n";
}