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 K = 1 << 6;
const int N = 1 << 12;
const int S = 1 << 8;
const int Q = 1 << 16;
const int M = 2.2e6;
int Arr[N], Pos[N], No[N][N][2], no;
int Mini[N / S][N + N], Maks[N / S][N + N], Ans[Q];
pair<int, int> Out[M];
int read(int *T, int l, int r, int ret, int (*best)(int, int)) {
  l += N;
  r += N;
  while (l < r) {
    ret = best(ret, T[l]);
    ret = best(ret, T[r]);
    l = (l + 1) / 2;
    r = (r - 1) / 2;
  }
  if (l == r) ret = best(ret, T[l]);
  return ret;
}
void calc(int l, int r, int cnt) {
  if (l == r) {
    No[l][r][cnt] = Pos[l];
    ;
    ;
    return;
  }
  int mid = (l + r) / 2;
  calc(l, mid, cnt ^ 1);
  calc(mid + 1, r, cnt ^ 1);
  vector<int> P;
  for (int i = l; i < r + 1; i++) P.push_back(Pos[i]);
  sort(P.begin(), P.end());
  ;
  for (auto a = P.begin(); a != P.end(); a++) {
    int lb = -1, le = -1, ub = -1, ue = -1;
    for (auto b = a; b != P.end(); b++) {
      if (Arr[*b] > mid) {
        if (ub == -1) ub = Arr[*b];
        ue = Arr[*b];
      } else {
        if (lb == -1) lb = Arr[*b];
        le = Arr[*b];
      }
      int aa = Arr[*a];
      int ab = Arr[*b];
      if (le == -1)
        No[aa][ab][cnt] = No[ub][ue][cnt ^ 1];
      else if (ue == -1)
        No[aa][ab][cnt] = No[lb][le][cnt ^ 1];
      else {
        Out[no] = {No[lb][le][cnt ^ 1], No[ub][ue][cnt ^ 1]};
        ;
        No[aa][ab][cnt] = no++;
      };
    }
  }
}
int mini(int a, int b) { return min(a, b); };
int maks(int a, int b) { return max(a, b); };
int main() {
  for (int j = 0; j < N / S; j++)
    for (int i = 0; i < N + N; i++) {
      Mini[j][i] = N;
      Maks[j][i] = -1;
    }
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &Arr[i]);
    Arr[i]--;
    Pos[Arr[i]] = i;
  }
  no = n;
  for (int l = 0; l < n; l += S) {
    int r = min(n, l + S) - 1;
    calc(l, r, 0);
  };
  for (int i = 0; i < n; i++) {
    int b = Arr[i] / S;
    Mini[b][N + i] = Maks[b][N + i] = i;
  }
  for (int b = 0; b < N / S; b++) {
    for (int i = N - 1; i > 0; i--) {
      Mini[b][i] = min(Mini[b][i * 2], Mini[b][i * 2 + 1]);
      Maks[b][i] = max(Maks[b][i * 2], Maks[b][i * 2 + 1]);
    }
  };
  for (int w = 0; w < q; w++) {
    int l, r;
    scanf("%d %d", &l, &r);
    l--;
    r--;
    int last = -1;
    ;
    for (int i = 0; i < N / S; i++) {
      int b = read(Mini[i], l, r, N, &mini);
      int e = read(Maks[i], l, r, 0, &maks);
      ;
      if (b == N) continue;
      b = Arr[b];
      e = Arr[e];
      if (last == -1)
        last = No[b][e][0];
      else {
        Out[no] = {last, No[b][e][0]};
        last = no++;
      }
    }
    Ans[w] = last;
  }
  printf("%d\n", no);
  for (int i = n; i < no; i++)
    printf("%d %d\n", Out[i].first + 1, Out[i].second + 1);
  for (int i = 0; i < q; i++) printf("%d ", Ans[i] + 1);
  printf("\n");
}