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 N = 1 << 12 | 5, Q = 1 << 16 | 5;
int n, q, B, a[N], p[N], cnt, id[Q];
vector<pair<int, int>> opt;
struct node {
  int n;
  vector<int> pos;
  vector<vector<int>> id;
  node(int _n = 0) {
    n = _n;
    pos.resize(n);
    id.resize(n);
    for (auto &v : id) {
      v.resize(n);
    }
  }
  inline int get(int l, int r) {
    l = lower_bound(pos.begin(), pos.end(), l) - pos.begin();
    r = upper_bound(pos.begin(), pos.end(), r) - pos.begin() - 1;
    return l <= r ? id[l][r] : 0;
  }
} sqr[N];
inline int merge(int x, int y) {
  int res;
  if (x && y) {
    opt.push_back({x, y});
    res = ++cnt;
  } else {
    res = x | y;
  }
  return res;
}
node solve(int l, int r, int s) {
  if (l == r) {
    node res(1);
    res.pos = {p[s + l]};
    res.id = {{p[s + l]}};
    return res;
  }
  int mid = l + r >> 1;
  node x = solve(l, mid, s);
  node y = solve(mid + 1, r, s);
  node z(x.n + y.n);
  merge(x.pos.begin(), x.pos.end(), y.pos.begin(), y.pos.end(), z.pos.begin());
  for (int i = 0; i < z.n; ++i) {
    for (int j = i; j < z.n; ++j) {
      int l = z.pos[i], r = z.pos[j];
      z.id[i][j] = merge(x.get(l, r), y.get(l, r));
    }
  }
  return z;
}
int main() {
  scanf("%d%d", &n, &q);
  B = sqrt(q);
  cnt = n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    p[a[i]] = i;
  }
  for (int i = 1, j = 1; i <= n; i += B, ++j) {
    sqr[j] = solve(0, min(i + B - 1, n) - i, i);
  }
  for (int t = 1; t <= q; ++t) {
    int l, r;
    scanf("%d%d", &l, &r);
    for (int i = 1, j = 1; i <= n; i += B, ++j) {
      id[t] = merge(id[t], sqr[j].get(l, r));
    }
  }
  printf("%d\n", cnt);
  for (auto x : opt) {
    printf("%d %d\n", x.first, x.second);
  }
  for (int t = 1; t <= q; ++t) {
    printf("%d%c", id[t], " \n"[t == q]);
  }
  return 0;
}