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>
int n, q, a[5001], l[200001], r[200001], cnt;
int set[51][301][301], top, val[301][301][301], fin[1000001];
int mn[301][5001];
std::vector<std::pair<int, int> > ans;
int merge(int a, int b) {
  ans.push_back(std::make_pair(a, b));
  return ++cnt;
}
void solve(std::vector<std::pair<int, int> >& a, int vl, int vr,
           int now[301][301]) {
  if (a.size() == 1) {
    now[0][0] = a[0].first;
    return;
  }
  if (!a.size()) return;
  std::vector<std::pair<int, int> > L, R;
  int vmid = (vl + vr) >> 1;
  for (int i = 0; i < a.size(); i++)
    if (a[i].second <= vmid)
      L.push_back(a[i]);
    else
      R.push_back(a[i]);
  ++top;
  solve(L, vl, vmid, set[top]);
  ++top;
  solve(R, vmid + 1, vr, set[top]);
  for (int i = 0; i < a.size(); i++)
    for (int j = i; j < a.size(); j++) {
      int l = std::lower_bound(L.begin(), L.end(),
                               std::make_pair(a[i].first, 0)) -
              L.begin(),
          r = std::lower_bound(L.begin(), L.end(),
                               std::make_pair(a[j].first + 1, 0)) -
              L.begin() - 1,
          x = 0;
      if (l <= r) x = set[top - 1][l][r];
      l = std::lower_bound(R.begin(), R.end(), std::make_pair(a[i].first, 0)) -
          R.begin(),
      r = std::lower_bound(R.begin(), R.end(),
                           std::make_pair(a[j].first + 1, 0)) -
          R.begin() - 1;
      if (l <= r && x)
        x = merge(x, set[top][l][r]);
      else
        x |= set[top][l][r];
      now[i][j] = x;
    }
  top -= 2;
}
std::vector<std::pair<int, int> > vec[5001];
int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
void write(int x) {
  if (x > 9) write(x / 10);
  putchar((x % 10) + '0');
}
int main() {
  n = read(), q = read();
  cnt = n;
  for (int i = 1; i <= n; i++) a[i] = read();
  int block = 256;
  for (int i = 1; i <= n; i++)
    vec[(a[i] - 1) / block + 1].push_back(std::make_pair(i, a[i]));
  for (int i = 1; i <= (n - 1) / block + 1; i++) {
    solve(vec[i], (i - 1) * block + 1, std::min(i * block, n), val[i]);
    for (int j = 1; j <= n + 1; j++) {
      mn[i][j] =
          std::lower_bound(vec[i].begin(), vec[i].end(), std::make_pair(j, 0)) -
          vec[i].begin();
    }
  }
  for (int i = 1, l, r; i <= q; i++) {
    l = read(), r = read();
    int x = 0;
    for (int j = 1; j <= (n - 1) / block + 1; j++) {
      int L = mn[j][l], R = mn[j][r + 1] - 1;
      if (L <= R && x)
        x = merge(x, val[j][L][R]);
      else
        x |= val[j][L][R];
    }
    fin[i] = x;
  }
  write(cnt);
  putchar('\n');
  for (int i = 0; i < ans.size(); i++)
    write(ans[i].first), putchar(' '), write(ans[i].second), putchar('\n');
  for (int i = 1; i <= q; i++) write(fin[i]), putchar(' ');
}