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, M = 1 << 16, B = 1 << 8, K = 1 << 22;
int n, m, idx, a[N], c[K], d[K], pos[N], ans[M];
int merge(int x, int y) {
  if (!x || !y) return x + y;
  c[idx] = x;
  d[idx] = y;
  return ++idx;
}
struct Seg {
  vector<int> val;
  vector<vector<int>> id;
  void init(int x) {
    val.resize(x);
    id.resize(x);
    for (int i = 0; i < x; i++) id[i].resize(x - i);
  }
  Seg(int x = -1) {
    if (x != -1) init(1), val[0] = pos[x], id[0][0] = pos[x] + 1;
  }
  int query(int l, int r) const {
    if (r < val.front() || l > val.back()) return 0;
    l = lower_bound(val.begin(), val.end(), l) - val.begin();
    r = upper_bound(val.begin(), val.end(), r) - val.begin() - 1;
    return l > r ? 0 : id[l][r - l];
  }
  Seg modify(const Seg &x, const Seg &y) {
    init(x.val.size() + y.val.size());
    merge(x.val.begin(), x.val.end(), y.val.begin(), y.val.end(), val.begin());
    for (int i = 0; i < int(val.size()); i++)
      for (int j = i; j < int(val.size()); j++)
        id[i][j - i] = merge(x.query(val[i], val[j]), y.query(val[i], val[j]));
    return *this;
  }
} t[N / B];
Seg solve(int l, int r) {
  if (l == r) return Seg(l);
  int mid = (l + r) >> 1;
  return Seg().modify(solve(l, mid), solve(mid + 1, r));
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  idx = n--;
  for (int i = 0; i <= n; i++) cin >> a[i], --a[i];
  for (int i = 0; i <= n; i++) pos[a[i]] = i;
  for (int i = 0; i <= n / B; i++) t[i] = solve(i * B, min(i * B + B - 1, n));
  int l, r;
  for (int i = 0; i < m; i++) {
    cin >> l >> r;
    l--;
    r--;
    for (int j = 0; j <= n / B; j++) ans[i] = merge(ans[i], t[j].query(l, r));
  }
  cout << idx << endl;
  for (int i = n + 1; i < idx; i++) cout << c[i] << ' ' << d[i] << '\n';
  for (int i = 0; i < m; i++) cout << ans[i] << ' ';
  return 0;
}