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, sq, cnt;
int a[4097], p[4097], bl[4097], ch[2200005][2];
vector<int> ve[4097];
vector<vector<int>> g[4097];
inline int merge(int x, int y) {
  if (!x || !y) return x | y;
  return ++cnt, ch[cnt][0] = x, ch[cnt][1] = y, cnt;
}
vector<vector<int>> solve(int l, int r) {
  if (l == r) {
    vector<vector<int>> f;
    f.resize(1);
    f[0].resize(1);
    return f[0][0] = p[l], f;
  }
  const int mid = l + r >> 1;
  vector<int> q, lq, rq;
  vector<vector<int>> L = solve(l, mid), R = solve(mid + 1, r), f;
  f.resize(r - l + 1);
  for (int i = 0; i < f.size(); ++i) f[i].resize(f.size());
  for (int i = l; i <= r; ++i) q.push_back(p[i]);
  sort(q.begin(), q.end());
  for (int i = l; i <= mid; ++i) lq.push_back(p[i]);
  sort(lq.begin(), lq.end());
  for (int i = mid + 1; i <= r; ++i) rq.push_back(p[i]);
  sort(rq.begin(), rq.end());
  for (int i = 0; i < q.size(); ++i)
    for (int j = i; j < q.size(); ++j) {
      int a = lower_bound(lq.begin(), lq.end(), q[i]) - lq.begin(),
          b = upper_bound(lq.begin(), lq.end(), q[j]) - lq.begin() - 1;
      int c = lower_bound(rq.begin(), rq.end(), q[i]) - rq.begin(),
          d = upper_bound(rq.begin(), rq.end(), q[j]) - rq.begin() - 1;
      if (a == L.size() || b == -1)
        f[i][j] = R[c][d];
      else if (c == R.size() || d == -1)
        f[i][j] = L[a][b];
      else
        f[i][j] = merge(L[a][b], R[c][d]);
    }
  return f;
}
int main() {
  scanf("%d%d", &n, &q);
  sq = sqrt(q);
  cnt = n;
  vector<int> ans;
  for (int i = 1; i <= n; ++i)
    scanf("%d", &a[i]), p[a[i]] = i, bl[i] = (i - 1) / sq + 1;
  for (int i = 1; i <= n; ++i) ve[bl[i]].push_back(p[i]);
  for (int i = 1; i <= bl[n]; ++i)
    sort(ve[i].begin(), ve[i].end()),
        g[i] = solve((i - 1) * sq + 1, min(i * sq, n));
  for (int i = 1, l, r; i <= q; ++i) {
    scanf("%d%d", &l, &r);
    int now = 0;
    for (int j = 1; j <= bl[n]; ++j) {
      int a = lower_bound(ve[j].begin(), ve[j].end(), l) - ve[j].begin(),
          b = upper_bound(ve[j].begin(), ve[j].end(), r) - ve[j].begin() - 1;
      if (a != ve[j].size() && b != -1) now = merge(now, g[j][a][b]);
    }
    ans.push_back(now);
  }
  printf("%d\n", cnt);
  for (int i = n + 1; i <= cnt; ++i) printf("%d %d\n", ch[i][0], ch[i][1]);
  for (int i = 0; i < ans.size(); ++i) printf("%d ", ans[i]);
  return puts(""), 0;
}