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, q, cnt;
int U[2200010], V[2200010];
int create(int u, int v) {
  cnt++;
  U[cnt] = u;
  V[cnt] = v;
  return cnt;
}
int a[5010];
int pos[5010];
struct Block {
  int l, r;
  int len() { return r - l + 1; }
  int p[310];
  vector<int> Id[310];
  Block() {}
  Block(int ll, int rr) {
    l = ll;
    r = rr;
    for (int i = l; i <= r; i++) {
      p[i - l] = pos[i];
      Id[i - l].resize(r - l + 1);
    }
    sort(p, p + len());
  }
  int getid(int l, int r) {
    l = lower_bound(p, p + len(), l) - p;
    r = upper_bound(p, p + len(), r) - p - 1;
    if (l > r) return 0;
    return Id[l][r];
  }
} S[5010];
int N;
Block solve(int l, int r) {
  Block res(l, r);
  if (l == r) {
    res.Id[0][0] = pos[l];
    return res;
  }
  int mid = (l + r) >> 1;
  Block Bl = solve(l, mid);
  Block Br = solve(mid + 1, r);
  int l1 = 0, r1;
  int l2 = 0, r2;
  for (int i = 0; i < res.len(); i++) {
    r1 = l1;
    r2 = l2;
    for (int j = i; j < res.len(); j++) {
      if (a[res.p[j]] <= mid)
        r1++;
      else
        r2++;
      if (l1 == r1)
        res.Id[i][j] = Br.Id[l2][r2 - 1];
      else if (l2 == r2)
        res.Id[i][j] = Bl.Id[l1][r1 - 1];
      else
        res.Id[i][j] = create(Bl.Id[l1][r1 - 1], Br.Id[l2][r2 - 1]);
    }
    if (a[res.p[i]] <= mid)
      l1++;
    else
      l2++;
  }
  return res;
}
int ans[100010];
int main() {
  scanf("%d %d", &n, &q);
  cnt = n;
  int Sz = sqrt(q);
  N = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  while (N * Sz + 1 <= n) {
    N++;
    S[N] = solve((N - 1) * Sz + 1, min(N * Sz, n));
  }
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    int u = 0;
    for (int j = 1; j <= N; j++) {
      int v = S[j].getid(l, r);
      if (!u)
        u = v;
      else if (v)
        u = create(u, v);
    }
    ans[i] = u;
  }
  printf("%d\n", cnt);
  for (int i = n + 1; i <= cnt; i++) printf("%d %d\n", U[i], V[i]);
  for (int i = 1; i <= q; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}